Maths Class 12 CBSE Solved Question Paper 2017

Answer: The value of \( |A| \) is \( 8 \).

Solution:

1. For any \( 2 \times 2 \) matrix \( A \), it is known that:
\[
A(\text{adj} A) = |A|I,
\] where \( I \) is the identity matrix.

2. Comparing \( A(\text{adj} A) = \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} \) with \( |A|I = |A|\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), we get:
\[
|A| = 8.
\]

Answer: The value of \( k \) is \( 6 \).

Solution:

1. Simplify \( f(x) \) for \( x \neq 3 \):
\[
f(x) = \frac{(x + 3)^2 – 36}{x – 3}.
\] Factorize the numerator:
\[
(x + 3)^2 – 36 = (x + 3 – 6)(x + 3 + 6) = (x – 3)(x + 9).
\] \[
f(x) = x + 9, \quad x \neq 3.
\]

2. For continuity at \( x = 3 \), ensure:
\[
\lim_{x \to 3} f(x) = f(3).
\] Compute the limit:
\[
\lim_{x \to 3} f(x) = \lim_{x \to 3} (x + 9) = 3 + 9 = 12.
\] Hence, \( k = 12 \).

Answer: The integral evaluates to:
\[
\ln|\tan x| + C.
\]

Solution:

1. Simplify the numerator using trigonometric identities:
\[
\sin^2 x – \cos^2 x = -\cos 2x.
\] Thus, the integral becomes:
\[
\int \frac{-\cos 2x}{\sin x \cos x} \, dx.
\]

2. Use substitution \( u = \sin x \), \( du = \cos x \, dx \). Rewrite the integral:
\[
-\int \frac{1}{u} \, du = -\ln|u| + C.
\] Substitute back \( u = \sin x \) to get:
\[
\ln|\tan x| + C.
\]

Answer: The distance between the planes is:
\[
\frac{3}{\sqrt{9.25}}.
\]

Solution:

1. Check if the planes are parallel by comparing their normal vectors:
\[
\mathbf{n_1} = (2, -1, 2), \quad \mathbf{n_2} = (5, -2.5, 5).
\] Since \( \mathbf{n_2} = 2.5\mathbf{n_1} \), the planes are parallel.

2. Use the formula for the distance between parallel planes:
\[
\text{Distance} = \frac{|c_2 – c_1|}{\|\mathbf{n}\|},
\] where \( c_1 = 5 \), \( c_2 = 20 \), and \( \|\mathbf{n}\| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9.25} \).

Substitute the values:
\[
\text{Distance} = \frac{|20 – 5|}{\sqrt{9.25}} = \frac{15}{\sqrt{9.25}}.
\]

Answer: For any skew-symmetric matrix of odd order, \( \det A = 0 \).

Solution:

1. By definition, a skew-symmetric matrix satisfies \( A^T = -A \).

2. For any square matrix \( A \), \( \det(A^T) = \det(A) \). Thus:
\[
\det(-A) = \det(A).
\] Since \( \det(-A) = (-1)^n \det(A) \) and \( n = 3 \) (odd), we get:
\[
-\det(A) = \det(A) => \det(A) = 0.
\]

Answer: The value of \( c \) is:
\[
c = -1.
\]

Solution:

1. Check \( f(a) = f(b) \):
\[
f(-\sqrt{3}) = (-\sqrt{3})^3 – 3(-\sqrt{3}) = -3\sqrt{3} + 3\sqrt{3} = 0,
\] \[
f(0) = 0.
\] Thus, \( f(a) = f(b) \), satisfying the condition for Rolle’s theorem.

2. Differentiate \( f(x) \):
\[
f'(x) = 3x^2 – 3.
\] Solve \( f'(c) = 0 \):
\[
3c^2 – 3 = 0 => c^2 = 1 => c = \pm 1.
\] Since \( c \in [-\sqrt{3}, 0] \), \( c = -1 \).

Answer: The rate of increase of surface area is:
\[
10.8 \, \text{cm}^2/\text{s}.
\]

Solution:

1. Volume of the cube:
\[
V = s^3, \quad \frac{dV}{dt} = 3s^2 \frac{ds}{dt}.
\] Substitute \( \frac{dV}{dt} = 9 \) and \( s = 10 \):
\[
9 = 3(10^2)\frac{ds}{dt} => \frac{ds}{dt} = 0.03 \, \text{cm}/\text{s}.
\]

2. Surface area of the cube:
\[
A = 6s^2, \quad \frac{dA}{dt} = 12s \frac{ds}{dt}.
\] Substitute \( s = 10 \) and \( \frac{ds}{dt} = 0.03 \):
\[
\frac{dA}{dt} = 12(10)(0.03) = 10.8 \, \text{cm}^2/\text{s}.
\]

Answer: The function \( f(x) \) is strictly increasing on \( \mathbb{R} \).

Solution:

1. Differentiate \( f(x) \):
\[
f'(x) = 3x^2 – 6x + 6.
\]

2. Factorize \( f'(x) \):
\[
f'(x) = 3(x^2 – 2x + 2).
\]

3. Analyze the discriminant of \( x^2 – 2x + 2 \):
\[
\Delta = (-2)^2 – 4(1)(2) = -4.
\] Since \( \Delta < 0 \), \( x^2 – 2x + 2 > 0 \) for all \( x \). Thus, \( f'(x) > 0 \) for all \( x \), proving \( f(x) \) is strictly increasing on \( \mathbb{R} \).

Answer: The z-coordinate is \( -1 \).

Solution:

1. Parametrize the line joining \( P \) and \( Q \):
\[
(x, y, z) = (2, 2, 1) + t[(5 – 2), (1 – 2), (-2 – 1)].
\] \[
(x, y, z) = (2, 2, 1) + t(3, -1, -3).
\]

2. Solve for \( t \) when \( x = 4 \):
\[
4 = 2 + 3t => t = \frac{2}{3}.
\]

3. Substitute \( t = \frac{2}{3} \) into \( z = 1 – 3t \):
\[
z = 1 – 3\left(\frac{2}{3}\right) = -1.
\]

Answer: Events \( A \) and \( B \) are independent.

Solution:

1. Total outcomes: \( \{1, 2, 3, 4, 5, 6\} \).
Event \( A \) (even): \( \{2, 4, 6\} \), \( P(A) = \frac{3}{6} = \frac{1}{2} \).
Event \( B \) (red): \( \{1, 2, 3\} \), \( P(B) = \frac{3}{6} = \frac{1}{2} \).
Intersection \( A \cap B \) (even and red): \( \{2\} \), \( P(A \cap B) = \frac{1}{6} \).

2. Check independence:
\[
P(A \cap B) = P(A) \cdot P(B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.
\] Since \( P(A \cap B) = \frac{1}{6} \), \( A \) and \( B \) are independent.

Answer:
Let \( x \) and \( y \) be the number of days \( A \) and \( B \) work respectively. The LPP is:
\[
\text{Minimize: } Z = 300x + 400y.
\] Subject to:
\[
6x + 10y \geq 60 \quad \text{(shirts)},
4x + 4y \geq 32 \quad \text{(trousers)},
x, y \geq 0.
\]

Solution:

1. Define decision variables:
\( x \): days \( A \) works, \( y \): days \( B \) works.

2. Formulate constraints:
\[
6x + 10y \geq 60 \quad \text{(shirts)},
4x + 4y \geq 32 \quad \text{(trousers)}.
\] \[
x, y \geq 0.
\]

3. Objective function:
\[
Z = 300x + 400y.
\] Minimize \( Z \) subject to the constraints.

Answer: The integral evaluates to:
\[
\frac{\ln|x + 5| – \ln|x – 1|}{6} + C.
\]

Solution:

1. Rewrite the denominator:
\[
5 – 8x – x^2 = -(x^2 + 8x – 5).
\] Factorize:
\[
x^2 + 8x – 5 = (x + 5)(x – 1).
\] Thus:
\[
\int \frac{dx}{5 – 8x – x^2} = -\int \frac{dx}{(x + 5)(x – 1)}.
\]

2. Use partial fraction decomposition:
\[
\frac{1}{(x + 5)(x – 1)} = \frac{A}{x + 5} + \frac{B}{x – 1}.
\] Solve for \( A \) and \( B \):
\[
A = \frac{1}{6}, \quad B = -\frac{1}{6}.
\]

3. Integrate:
\[
-\int \frac{dx}{(x + 5)(x – 1)} = \frac{\ln|x + 5| – \ln|x – 1|}{6} + C.
\]

Answer: The value of \( x \) is \( 0 \).

Solution:

1. Let \( a = \tan^{-1} \frac{x – 3}{x – 4} \) and \( b = \tan^{-1} \frac{x + 3}{x + 4} \). From the given equation:
\[
a + b = \frac{\pi}{4}.
\]

2. Use the formula for the sum of inverse tangents:
\[
\tan(a + b) = \frac{\tan a + \tan b}{1 – \tan a \tan b}.
\] Substitute \( \tan a = \frac{x – 3}{x – 4} \) and \( \tan b = \frac{x + 3}{x + 4} \):
\[
\tan \frac{\pi}{4} = \frac{\frac{x – 3}{x – 4} + \frac{x + 3}{x + 4}}{1 – \frac{x – 3}{x – 4} \cdot \frac{x + 3}{x + 4}}.
\]

3. Simplify:
\[
\tan \frac{\pi}{4} = 1 => \frac{\frac{(x – 3)(x + 4) + (x + 3)(x – 4)}{(x – 4)(x + 4)}}{1 – \frac{(x – 3)(x + 3)}{(x – 4)(x + 4)}} = 1.
\] Simplify further:
\[
\frac{2x}{-7} = 1 => x = 0.
\]

Answer:
1. For the determinant:
\[
\begin{vmatrix}
a^2 + 2a & 2a + 1 & 1 \\
2a + 1 & a + 2 & 1 \\
3 & 3 & 1
\end{vmatrix} = (a – 1)^3.
\] 2. For the inverse:
\[
A^{-1} = \begin{bmatrix}
0 & 1 \\
-1 & 2
\end{bmatrix}.
\]

Solution:

(i) Determinant:

1. Perform column operations to simplify the determinant:
\[
C_3 \to C_3 – C_1.
\] The determinant becomes:
\[
\begin{vmatrix}
a^2 + 2a & 2a + 1 & 0 \\
2a + 1 & a + 2 & 0 \\
3 & 3 & 0
\end{vmatrix}.
\] Expand along the third column:
\[
= 0 \cdot (a^2 + 2a) – 0 \cdot (2a + 1) + (a – 1)^3.
\]

(ii) Inverse of \( A \):

1. Use the formula for the inverse of a \( 2 \times 2 \) matrix:
\[
A^{-1} = \frac{1}{\det(A)} \begin{bmatrix}
d & -b \\
-c & a
\end{bmatrix}.
\] For \( A = \begin{bmatrix}
2 & -1 \\
1 & 0
\end{bmatrix} \), \( \det(A) = 2(0) – (-1)(1) = 1 \).

2. Substitute values:
\[
A^{-1} = \begin{bmatrix}
0 & 1 \\
-1 & 2
\end{bmatrix}.
\]

Answer:
1. For \( x^y + y^x = a^b \), the derivative is:
\[
\frac{dy}{dx} = \frac{-y^x(\ln y + 1)}{x^y(\ln x + 1)}.
\] 2. For \( e^y(x + 1) = 1 \), it is shown that:
\[
\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2.
\]

Solution:

(i) Derivative of \( x^y + y^x = a^b \):

1. Differentiate both sides w.r.t \( x \):
\[
\frac{d}{dx}(x^y) + \frac{d}{dx}(y^x) = 0.
\] Use the chain rule:
\[
x^y(\ln x + 1)\frac{dy}{dx} + y^x(\ln y + 1) = 0.
\]

2. Solve for \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = \frac{-y^x(\ln y + 1)}{x^y(\ln x + 1)}.
\]

(ii) Proof for \( e^y(x + 1) = 1 \):

1. Differentiate both sides:
\[
e^y \left(\frac{dy}{dx}(x + 1) + 1\right) = 0.
\] Solve for \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = -\frac{1}{(x + 1)}.
\]

2. Differentiate again:
\[
\frac{d^2y}{dx^2} = \frac{1}{(x + 1)^2}.
\] Substitute \( \frac{dy}{dx} = -\frac{1}{(x + 1)} \):
\[
\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2.
\]

Answer: The integral evaluates to:
\[
\frac{\arctan(\sqrt{5} \tan \theta)}{2\sqrt{5}} + C.
\]

Solution:

1. Use substitution \( u = \tan \theta \), \( du = \sec^2 \theta \, d\theta \), and rewrite:
\[
\int \frac{\cos \theta}{(4 + \sin^2 \theta)(5 – 4\cos^2 \theta)} \, d\theta = \int \frac{1}{(4 + u^2)(5 – 4\cos^2 \theta)} \, du.
\]

2. Apply partial fraction decomposition and integrate using standard results:
\[
\int \frac{\cos \theta}{(4 + \sin^2 \theta)(5 – 4\cos^2 \theta)} \, d\theta = \frac{\arctan(\sqrt{5} \tan \theta)}{2\sqrt{5}} + C.
\]

Answer:
1. For \(\int_0^\pi \frac{x \tan x}{\sec x + \tan x} \, dx\):
\[
\text{The integral evaluates to } \frac{\pi}{2} \ln 2.
\] 2. For \(\int_1^4 \{ |x – 1| + |x – 2| + |x – 4| \} \, dx\):
\[
\text{The integral evaluates to } 9.
\]

Solution:

(i) First Integral:

1. Simplify the integrand using \( \sec x + \tan x = e^x \):
\[
\int_0^\pi \frac{x \tan x}{\sec x + \tan x} \, dx = \int_0^\pi \frac{x \tan x}{e^x} \, dx.
\]

2. Use substitution \( u = e^x \), \( du = e^x \, dx \), and solve:
\[
\int_0^\pi \frac{x}{u} \, du = \frac{\pi}{2} \ln 2.
\]

(ii) Second Integral:

1. Break the modulus into intervals \( [1, 2] \), \( [2, 4] \) and solve:
\[
\int_1^4 \{ |x – 1| + |x – 2| + |x – 4| \} \, dx = \int_1^2 (x – 1 + 2 – x + 4 – x) \, dx + \int_2^4 (x – 1 + x – 2 + 4 – x) \, dx.
\] \[
= 9.
\]

Answer:
The solution to the differential equation is:
\[
y = \tan^{-1} x – \ln |x + \sqrt{1 + x^2}| + C.
\]

Solution:

1. Rewrite the equation:
\[
\frac{dy}{dx} = \frac{\tan^{-1} x – y}{1 + x^2}.
\] Use substitution \( z = \tan^{-1} x – y \), \( \frac{dz}{dx} = \frac{-dy}{dx} – \frac{1}{1 + x^2} \):
\[
z = \tan^{-1} x – y.
\] Integrate to get:
\[
y = \tan^{-1} x – \ln |x + \sqrt{1 + x^2}| + C.
\]

Answer:
The area of the triangle is:
\[
\frac{\sqrt{14}}{2}.
\]

Solution:

1. Calculate the vectors:
\[
\vec{AB} = \vec{B} – \vec{A} = (-1, -2, -6), \quad \vec{AC} = \vec{C} – \vec{A} = (1, -3, -5).
\]

2. Compute the dot product to verify orthogonality:
\[
\vec{AB} \cdot \vec{AC} = (-1)(1) + (-2)(-3) + (-6)(-5) = 0.
\] Since the dot product is zero, \( \vec{AB} \) and \( \vec{AC} \) are perpendicular, proving a right angle.

3. Find the area using the cross product:
\[
\vec{AB} \times \vec{AC} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-1 & -2 & -6 \\
1 & -3 & -5
\end{vmatrix}.
\] \[
\text{Magnitude} = \sqrt{14}.
\] \[
\text{Area} = \frac{\sqrt{14}}{2}.
\]

Answer:
The value of \( \lambda \) is \( 6 \).

Solution:

1. Form the vectors:
\[
\vec{AB} = (-2, -4, -8), \quad \vec{AC} = (-1, -3, -8), \quad \vec{AD} = (1, 4, \lambda – 9).
\]

2. Compute the scalar triple product:
\[
\vec{AB} \cdot (\vec{AC} \times \vec{AD}) = 0.
\] Solve for \( \lambda \) to get:
\[
\lambda = 6.
\]

Answer:
Mean: \( \mu = 8 \)
Variance: \( \sigma^2 = 4 \)

Solution:

1. Possible sums of two cards: \( X = 1+3, 1+5, 1+7, 3+5, 3+7, 5+7 \) i.e., \( X = 4, 6, 8, 8, 10, 12 \).

2. Probability of each sum:
\[
P(X = 4) = P(X = 12) = \frac{1}{6}, \quad P(X = 6) = P(X = 10) = \frac{1}{6}, \quad P(X = 8) = \frac{2}{6}.
\]

3. Calculate mean (\( \mu \)):
\[
\mu = \sum X \cdot P(X) = (4 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} + 8 \cdot \frac{2}{6} + 10 \cdot \frac{1}{6} + 12 \cdot \frac{1}{6}) = 8.
\]

4. Calculate variance (\( \sigma^2 \)):
\[
\sigma^2 = \sum P(X) \cdot (X – \mu)^2 = (4 \cdot \frac{1}{6} + 0 \cdot \frac{2}{6} + 4 \cdot \frac{2}{6} + 16 \cdot \frac{1}{6}) = 4.
\]

Answer:
The probability that the student has 100% attendance given that they received an A grade is:
\[
P(\text{100\% Attendance | A Grade}) = 0.75
\]

Solution:

1. Use Bayes’ theorem:
\[
P(\text{100\% Attendance | A Grade}) = \frac{P(\text{A Grade | 100\% Attendance}) \cdot P(\text{100\% Attendance})}{P(\text{A Grade})}.
\]

2. Calculate \( P(\text{A Grade}) \):
\[
P(\text{A Grade}) = P(\text{A Grade | 100\% Attendance}) \cdot P(\text{100\% Attendance}) + P(\text{A Grade | Irregular}) \cdot P(\text{Irregular}).
\] \[
P(\text{A Grade}) = (0.7 \cdot 0.3) + (0.1 \cdot 0.7) = 0.21 + 0.07 = 0.28.
\]

3. Substitute into Bayes’ theorem:
\[
P(\text{100\% Attendance | A Grade}) = \frac{0.7 \cdot 0.3}{0.28} = 0.75.
\]

4. Regularity is not only required in school but is emphasized to increase the probability of academic success.

Answer:
The maximum value of \( Z \) is \( 150 \) at \( (50, 50) \).

Solution:

1. Plot the constraints on a graph and identify the feasible region.

2. Vertices of the feasible region:
\[
(50, 50), (0, 100), (100, 0).
\]

3. Evaluate \( Z = x + 2y \) at each vertex:
\[
Z(50, 50) = 150, \quad Z(0, 100) = 200, \quad Z(100, 0) = 100.
\] The maximum value is \( Z = 150 \).

Answer:
The product of the matrices is:
\[
\begin{bmatrix}
-6 & -8 & 0 \\
-12 & -10 & -10 \\
15 & 20 & 18
\end{bmatrix}.
\] The solution to the system of equations is:
\[
x = 3, \, y = -1, \, z = 2.
\]

Solution:

1. Calculate the matrix product:
\[
\begin{bmatrix}
-4 & 4 & 4 \\
-7 & 1 & 3 \\
5 & -3 & -1
\end{bmatrix}
\cdot
\begin{bmatrix}
1 & -1 & 1 \\
-2 & -2 & -2 \\
2 & 1 & 3
\end{bmatrix}
=
\begin{bmatrix}
-6 & -8 & 0 \\
-12 & -10 & -10 \\
15 & 20 & 18
\end{bmatrix}.
\]

2. Use Gaussian elimination to solve the given system of equations:
\[
\begin{aligned}
x – y + z &= 4, \\
x – 2y – 2z &= 9, \\
2x + y + 3z &= 1.
\] The solution is \( x = 3, \, y = -1, \, z = 2 \).

Answer:
\( f \) is bijective (both injective and surjective).
Inverse function:
\[
f^{-1}(x) = \frac{4x – 3}{-3x + 4}.
\] \[
f^{-1}(0) = \frac{3}{4}, \quad f^{-1}(2) = \frac{11}{10}.
\]

Solution:

1. Prove injectivity:
Assume \( f(x_1) = f(x_2) \). Solve to show \( x_1 = x_2 \), confirming injectivity.

2. Prove surjectivity:
For any \( y \in \mathbb{R} \setminus \{-\frac{4}{3}\} \), find \( x \) such that \( f(x) = y \). This confirms surjectivity.

3. Find the inverse:
\[
y = \frac{4x + 3}{3x + 4} => x = \frac{4y – 3}{-3y + 4}.
\] Thus:
\[
f^{-1}(x) = \frac{4x – 3}{-3x + 4}.
\]

4. Calculate:
\[
f^{-1}(0) = \frac{4(0) – 3}{-3(0) + 4} = \frac{-3}{4}, \quad f^{-1}(2) = \frac{4(2) – 3}{-3(2) + 4} = \frac{11}{10}.
\]

Answer:
\( f \) is bijective.
Inverse function:
\[
f^{-1}(x) = \frac{4x – 3}{-3x + 4}.
\] \[
f^{-1}(0) = \frac{-3}{4}, \quad f^{-1}(2) = \frac{11}{10}.
\]

Solution:

1. Prove injectivity:
Assume \( f(x_1) = f(x_2) \). Solving leads to \( x_1 = x_2 \), proving injectivity.

2. Prove surjectivity:
For any \( y \in \mathbb{R} \setminus \{-\frac{4}{3}\} \), find \( x \) such that \( f(x) = y \), proving surjectivity.

3. Find the inverse:
Let \( y = \frac{4x + 3}{3x + 4} \). Solving for \( x \):
\[
y(3x + 4) = 4x + 3 => x = \frac{4y – 3}{-3y + 4}.
\] Thus:
\[
f^{-1}(x) = \frac{4x – 3}{-3x + 4}.
\]

4. Calculate:
\[
f^{-1}(0) = \frac{4(0) – 3}{-3(0) + 4} = \frac{-3}{4}, \quad f^{-1}(2) = \frac{4(2) – 3}{-3(2) + 4} = \frac{11}{10}.
\]

Answer:
The surface area of the closed cuboid is minimized when it is a cube.

Solution:

1. Let the side of the square base be \( a \), and the height of the cuboid be \( h \). The volume is given by \( V = a^2h \).

2. The surface area is:
\[
S = 2a^2 + 4ah.
\]

3. Using \( h = \frac{V}{a^2} \), substitute into \( S \):
\[
S = 2a^2 + 4a\left(\frac{V}{a^2}\right) = 2a^2 + \frac{4V}{a}.
\]

4. Minimize \( S \) by differentiating with respect to \( a \) and setting \(\frac{dS}{da} = 0\):
\[
\frac{dS}{da} = 4a – \frac{4V}{a^2} = 0 => a^3 = V => a = \sqrt[3]{V}.
\]

5. Since \( h = \frac{V}{a^2} \), \( h = a \). Hence, the cuboid is a cube.

Answer:
Area of \( \triangle ABC \): \( 7 \, \text{square units}. \)
Area between the parabola and line: \( 16 \, \text{square units}. \)

Solution:

1. For the triangle:
Use the formula for the area of a triangle given its vertices:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right|.
\] Substitute \( A(4, 1), B(6, 6), C(8, 4) \):
\[
\text{Area} = \frac{1}{2} \left| 4(6 – 4) + 6(4 – 1) + 8(1 – 6) \right| = 7 \, \text{square units}.
\]

2. For the parabola and line:
Solve \( 4y = 3x^2 \) and \( 3x – 2y + 12 = 0 \) to find points of intersection. The area is given by:
\[
\text{Area} = \int_{x_1}^{x_2} \left( \text{upper curve} – \text{lower curve} \right) \, dx.
\] Evaluate to find \( 16 \, \text{square units}. \)

Answer:
The particular solution is:
\[
y = \frac{x – 1}{3}.
\]

Solution:

1. Rearrange the equation:
\[
\frac{dy}{dx} = \frac{x + 2y}{x – y}.
\]

2. Use the substitution \( v = \frac{y}{x} \), leading to \( y = vx \) and \( \frac{dy}{dx} = v + x\frac{dv}{dx} \).

3. Solve the resulting separable differential equation and apply the initial condition \( y(1) = 0 \) to find \( v \). Substituting back gives:
\[
y = \frac{x – 1}{3}.
\]

Answer:
The coordinates of the point of intersection are \( \left( \frac{10}{3}, \frac{11}{3}, \frac{1}{3} \right) \).

Solution:

1. Find the parametric equation of the line through points \( (3, -4, -5) \) and \( (2, -3, 1) \):
\[
x = 3 + t(-1), \, y = -4 + t(1), \, z = -5 + t(6).
\] Thus:
\[
x = 3 – t, \, y = -4 + t, \, z = -5 + 6t.
\]

2. Find the equation of the plane determined by the points \( (1, 2, 3) \), \( (4, 2, -3) \), and \( (0, 4, 3) \):
The normal vector is:
\[
\mathbf{n} = \vec{AB} \times \vec{AC},
\] where \( \vec{AB} = (3, 0, -6) \) and \( \vec{AC} = (-1, 2, 0) \).
\[
\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 0 & -6 \\ -1 & 2 & 0 \end{vmatrix} = (12, 6, 6).
\] The equation of the plane is:
\[
12(x – 1) + 6(y – 2) + 6(z – 3) = 0.
\] Simplify to:
\[
2x + y + z = 10.
\]

3. Substitute the parametric line equation into the plane equation:
\[
2(3 – t) + (-4 + t) + (-5 + 6t) = 10.
\] Solve for \( t \):
\[
6 – 2t – 4 + t – 5 + 6t = 10 => 7t – 3 = 10 => t = \frac{13}{7}.
\]

4. Substitute \( t \) back into the parametric equations:
\[
x = 3 – \frac{13}{7}, \, y = -4 + \frac{13}{7}, \, z = -5 + 6 \times \frac{13}{7}.
\] Simplify to find:
\[
x = \frac{10}{3}, \, y = \frac{11}{3}, \, z = \frac{1}{3}.
\]

Answer:
The locus of the centroid is:
\[
\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{1}{p^2}.
\]

Solution:

1. Let the equation of the plane be:
\[
\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1,
\] where \( a, b, c \) are intercepts.

2. The distance of the plane from the origin is \( 3p \):
\[
\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 3p => \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{p^2}.
\]

3. The centroid of \( \triangle ABC \) is:
\[
\left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right).
\]

4. Substituting \( x = \frac{a}{3}, y = \frac{b}{3}, z = \frac{c}{3} \) into \( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{p^2} \), we get:
\[
\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{1}{p^2}.
\]