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Calculation:

Given data:

Direction ratios of the line parallel to the plane P are (1, 3, 2) .

The equation of the two planes are:

Plane 1: 6x + 4y - 5z = 2 , normal vector $\mathbf{n_1} = (6, 4, -5)$

Plane 2:  x - 2y + 3z = 0 , normal vector $\mathbf{n_2} = (1, -2, 3)$

The direction ratios of the line of intersection of two planes can be found using the cross product of the normal vectors of the planes:

$\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2}$

$\mathbf{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 6 & 4 & -5 \ 1 & -2 & 3 \end{vmatrix}$

Expanding the determinant:

$\mathbf{d} = \hat{i} \begin{vmatrix} 4 & -5 \ -2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 6 & -5 \ 1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 6 & 4 \ 1 & -2 \end{vmatrix}$

$\mathbf{d} = \hat{i} \left( (4)(3) - (-5)(-2) \right) - \hat{j} \left( (6)(3) - (-5)(1) \right) + \hat{k} \left( (6)(-2) - (4)(1) \right)$

$\mathbf{d} = \hat{i} (12 - 10) - \hat{j} (18 + 5) + \hat{k} (-12 - 4)$

$\mathbf{d} = \hat{i} (2) - \hat{j} (23) + \hat{k} (-16)$

Therefore, the direction ratios of the line of intersection are  (2, -23, -16) .

Hence, the Correct answer is Option 2.

Calculation:

Family of planes through the intersection

The general form is:

$(6x + 4y - 5z - 2) + \lambda(x - 2y + 3z) = 0$

Expanding:

$(6 + \lambda)x + (4 - 2\lambda)y + (-5 + 3\lambda)z - 2 = 0$

Apply parallel condition

Let normal vector $\vec{n} = \langle 6 + \lambda, 4 - 2\lambda, -5 + 3\lambda \rangle$

Direction vector of the line: $\vec{d} = \langle 1, 3, 2 \rangle$

Since the plane is parallel to the line, $\vec{n} \cdot \vec{d} = 0$

Compute dot product:

$(6 + \lambda)(1) + (4 - 2\lambda)(3) + (-5 + 3\lambda)(2) = 0$

$6 + \lambda + 12 - 6\lambda - 10 + 6\lambda = 0 \Rightarrow \lambda + 8 = 0 \Rightarrow \lambda = -8$

Substitute back to get plane equation

$(6 - 8)x + (4 + 16)y + (-5 - 24)z - 2 = 0 \Rightarrow -2x + 20y - 29z - 2 = 0$

Multiply by -1 to simplify:

$2x - 20y + 29z + 2 = 0$

Hence, the required plane equation is:

$2x - 20y + 29z + 2 = 0$

Calculation:

Let the centre of the sphere be $O(x,y,z)$.

Distances of the centre from the given points:

⇒ $OA=\sqrt{(x-1)^2+y^2+z^2}$

⇒ $OB=\sqrt{x^2+(y-1)^2+z^2}$

⇒ $OC=\sqrt{x^2+y^2+(z-1)^2}$

Since the sphere passes through all three points: $OA=OB$

⇒ $(x-1)^2+y^2+z^2=x^2+(y-1)^2+z^2$

⇒ $x^2-2x+1+y^2+z^2=x^2+y^2-2y+1+z^2$

$\Rightarrow x=y$

Also, $OB=OC$

⇒ $x^2+(y-1)^2+z^2=x^2+y^2+(z-1)^2$

⇒ $x^2+y^2-2y+1+z^2=x^2+y^2+z^2-2z+1$

$\Rightarrow y=z$

Hence, $x=y=z$

For the smallest radius:

⇒ $x=y=z=\frac{1}{3}$

Radius of the sphere:

⇒ $r=\sqrt{\left(1-\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2}$

$=\sqrt{\frac{4}{9}+\frac{1}{9}+\frac{1}{9}} =\sqrt{\frac{6}{9}} =\sqrt{\frac{2}{3}}$

Hence, the correct answer is Option 2

Calculation:

We are given three points: A(1, 0, 0),  B(0, 1, 0) , and  C(0, 0, 1) , and we are asked to find the equation of the plane passing through these points and the center of the sphere.

$\overrightarrow{AB} = B - A = (0 - 1, 1 - 0, 0 - 0) = (-1, 1, 0)$

$\overrightarrow{AC} = C - A = (0 - 1, 0 - 0, 1 - 0) = (-1, 0, 1)$

$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 1 & 0 \ -1 & 0 & 1 \end{vmatrix} = \hat{i}(1) - \hat{j}(-1) + \hat{k}(1) = \hat{i} + \hat{j} + \hat{k}$

Thus, the normal vector to the plane is (1, 1, 1).

The equation of the plane is $x + y + z - 1 = 0$

The center of the sphere lies on this plane. Therefore, the center of the sphere lies on the plane with the equation x + y + z - 1 = 0 

Hence, the correct answer is Option 1.

Calculation:

$A(1, -1, 0), B(-2, 1, 8), C(-1, 2, 7)$ are the three consecutive vertices of the parallelogram. Let the fourth vertex be $D(x, y, z)$.

In a parallelogram, the diagonals bisect each other. Therefore, the midpoint of diagonal $AC$ must be equal to the midpoint of diagonal $BD$.

Midpoint of AC$= \left( \frac{1 + (-1)}{2}, \frac{-1 + 2}{2}, \frac{0 + 7}{2} \right) = (0, \frac{1}{2}, \frac{7}{2})$

Midpoint of BD$= \left( \frac{-2 + x}{2}, \frac{1 + y}{2}, \frac{8 + z}{2} \right)$

Equating the midpoints, we get

$\frac{-2 + x}{2} = 0, \quad \frac{1 + y}{2} = \frac{1}{2}, \quad \frac{8 + z}{2} = \frac{7}{2}$

Solving for $x, y, z$:

$x = 2, \quad y = 0, \quad z = -1$

Therefore, the fourth vertex D is $(2, 0, -1)$.

Hence, the correct answer is Option 2.

Calculation:

$\vec{BC} = \hat{i} + \hat{j} - \hat{k}$

$\vec{CD} = 3\hat{i} - 2\hat{j} - 8\hat{k}$

$\cos \theta = \frac{\vec{BC} \cdot \vec{CD}}{|\vec{BC}| \cdot |\vec{CD}|}$

Compute dot product:

$\vec{BC} \cdot \vec{CD} = 1 \cdot 3 + 1 \cdot (-2) + (-1) \cdot (-8) = 3 - 2 + 8 = 9$

Compute magnitudes:

$|\vec{BC}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$

$|\vec{CD}| = \sqrt{3^2 + (-2)^2 + (-8)^2} = \sqrt{9 + 4 + 64} = \sqrt{77}$

$\cos \theta = \frac{9}{\sqrt{3} \cdot \sqrt{77}}$

$\cos^2 \theta = \left( \frac{9}{\sqrt{3} \cdot \sqrt{77}} \right)^2 = \frac{81}{3 \cdot 77} = \frac{27}{77}$

Hence, the correct answer is Option 2.

Calculation:

4y = mx-m+2

4y = m(x-1)+2

$y - \frac{1}{2} = \frac{m}{4}(x -1)$

for any values of m the family of lines always posses through (1, 1/2) so, all lines are concurrent.

Calculation:

Let locus of point be $(x, y)$

⇒ $(x - a)^2 + (y - b)^2 = (x - c)^2 + (y - d)^2$

⇒ $x^2 - 2ax + a^2 + y^2 - 2by + b^2 = x^2 - 2cx + c^2 + y^2 - 2dy + d^2$

⇒ $2(a - c)x + 2(b - d)y + c^2 + d^2 - a^2 - b^2 = 0$

⇒ $(a - c)x + (b - d)y + \frac{1}{2}(c^2 + d^2 - a^2 - b^2) = 0$

$\therefore \quad k = \frac{(c^2 + d^2 - a^2 - b^2)}{2}$

Hence, the correct answer is Option 4.

Calculation:

Given: The equation is $x^2 + 3y = 0$

Rewriting: $y = -\frac{x^2}{3}$, which is the standard form of a parabola.

Statement I: The equation represents a parabola that opens upwards.

Since $a = -\frac{1}{3}$ is negative, the parabola opens downwards. Therefore, Statement I is incorrect

Statement II: The axis of the parabola is $x = 0$

In the equation $x^2 + 3y = 0$, the variable x  is squared, and there is no linear term in x . Hence, the axis of symmetry is along the y -axis, i.e., x = 0 . Therefore, Statement II is correct

Statement III: The equation of the latus rectum is $4y - 3 = 0$

The latus rectum's length for a parabola $y = ax^2$ is$\frac{4}{|a|}$, where $a = -\frac{1}{3}$ so the length is $\frac{4}{\left| -\frac{1}{3} \right|} = 12$. The equation  4y - 3 = 0  does not represent the latus rectum equation. Therefore, Statement III is incorrect.

Only Statement II is correct.

Therefore, the correct answer is Option 2

Calculation:

Put $y = 0$, we get x-intercepts:

$x = \frac{2a^2}{a^2 + b^2}$

Put $x = 0$, we get y-intercepts:

$y = \frac{2b^2}{a^2 + b^2}$

$\text{Sum} = \frac{2a^2}{a^2 + b^2} + \frac{2b^2}{a^2 + b^2} = 2$

Hence, the correct answer is Option 1.