Plane Questions in English

(A) Let the given points be $A(1,1,0), B(1,2,1),$ and $C(-2,2,-1)$.
The equation of a plane passing through three points $(x_1, y_1, z_1), (x_2, y_2, z_2),$ and $(x_3, y_3, z_3)$ is given by the determinant equation:
$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$
Substituting the coordinates of points $A, B,$ and $C$:
$\begin{vmatrix} x-1 & y-1 & z-0 \\ 1-1 & 2-1 & 1-0 \\ -2-1 & 2-1 & -1-0 \end{vmatrix} = 0$
$\begin{vmatrix} x-1 & y-1 & z \\ 0 & 1 & 1 \\ -3 & 1 & -1 \end{vmatrix} = 0$
Expanding along the first row:
$(x-1)(-1-1) - (y-1)(0 - (-3)) + z(0 - (-3)) = 0$
$(x-1)(-2) - (y-1)(3) + z(3) = 0$
$-2x + 2 - 3y + 3 + 3z = 0$
$-2x - 3y + 3z + 5 = 0$
Multiplying by $-1$,we get:
$2x + 3y - 3z = 5$

(A) The given equation of the plane is $2x + y - z = 5$.
Dividing both sides of the equation by $5$,we get:
$\frac{2x}{5} + \frac{y}{5} - \frac{z}{5} = 1$
This can be rewritten in the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ as:
$\frac{x}{\frac{5}{2}} + \frac{y}{5} + \frac{z}{-5} = 1$
Comparing this with the standard intercept form,the intercepts $a, b, c$ are:
$a = \frac{5}{2}$,$b = 5$,and $c = -5$.
Thus,the intercepts cut off by the plane are $\frac{5}{2}, 5, -5$.

(A) The equation of the $ZOX$ plane is $y=0$.
Any plane parallel to the $ZOX$ plane is of the form $y=k$,where $k$ is a constant.
Given that the plane has an intercept of $3$ on the $y$-axis,the value of $k$ must be $3$.
Therefore,the equation of the required plane is $y=3$.

(A) The equation of any plane passing through the intersection of the planes $P_1: 3x - y + 2z - 4 = 0$ and $P_2: x + y + z - 2 = 0$ is given by $P_1 + \lambda P_2 = 0$.
$(3x - y + 2z - 4) + \lambda(x + y + z - 2) = 0$ --- $(1)$
Since the plane passes through the point $(2, 2, 1)$,this point must satisfy the equation $(1)$.
Substituting $x = 2, y = 2, z = 1$ in equation $(1)$:
$(3(2) - 2 + 2(1) - 4) + \lambda(2 + 2 + 1 - 2) = 0$
$(6 - 2 + 2 - 4) + \lambda(3) = 0$
$2 + 3\lambda = 0$
$\lambda = -\frac{2}{3}$
Substituting $\lambda = -\frac{2}{3}$ back into equation $(1)$:
$(3x - y + 2z - 4) - \frac{2}{3}(x + y + z - 2) = 0$
$3(3x - y + 2z - 4) - 2(x + y + z - 2) = 0$
$(9x - 3y + 6z - 12) - (2x + 2y + 2z - 4) = 0$
$7x - 5y + 4z - 8 = 0$
Thus,the required equation of the plane is $7x - 5y + 4z - 8 = 0$.

(A) The equations of the given planes are $\vec{r} \cdot (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 5$ and $\vec{r} \cdot (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) = 3$.
The normal vectors to the planes are $\vec{n}_{1} = 2 \hat{i} + 2 \hat{j} - 3 \hat{k}$ and $\vec{n}_{2} = 3 \hat{i} - 3 \hat{j} + 5 \hat{k}$.
The angle $Q$ between the two planes is given by the formula:
$\cos Q = \left| \frac{\vec{n}_{1} \cdot \vec{n}_{2}}{|\vec{n}_{1}| |\vec{n}_{2}|} \right|$
First,calculate the dot product $\vec{n}_{1} \cdot \vec{n}_{2}$:
$\vec{n}_{1} \cdot \vec{n}_{2} = (2)(3) + (2)(-3) + (-3)(5) = 6 - 6 - 15 = -15$.
Next,calculate the magnitudes $|\vec{n}_{1}|$ and $|\vec{n}_{2}|$:
$|\vec{n}_{1}| = \sqrt{2^2 + 2^2 + (-3)^2} = \sqrt{4 + 4 + 9} = \sqrt{17}$.
$|\vec{n}_{2}| = \sqrt{3^2 + (-3)^2 + 5^2} = \sqrt{9 + 9 + 25} = \sqrt{43}$.
Now,substitute these values into the formula:
$\cos Q = \left| \frac{-15}{\sqrt{17} \cdot \sqrt{43}} \right| = \frac{15}{\sqrt{731}}$.
Thus,the angle $Q$ is $\cos^{-1} \left( \frac{15}{\sqrt{731}} \right)$.

(A) The direction ratios of the normal to the planes $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0$ are $(A_1, B_1, C_1)$ and $(A_2, B_2, C_2)$ respectively.
The planes are parallel if $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$.
The planes are perpendicular if $A_1A_2 + B_1B_2 + C_1C_2 = 0$.
The angle $\theta$ between the planes is given by $\cos \theta = \left| \frac{A_1A_2 + B_1B_2 + C_1C_2}{\sqrt{A_1^2 + B_1^2 + C_1^2} \sqrt{A_2^2 + B_2^2 + C_2^2}} \right|$.
For the given planes $7x + 5y + 6z + 30 = 0$ and $3x - y - 10z + 4 = 0$:
$A_1 = 7, B_1 = 5, C_1 = 6$
$A_2 = 3, B_2 = -1, C_2 = -10$
Check for perpendicularity:
$A_1A_2 + B_1B_2 + C_1C_2 = (7)(3) + (5)(-1) + (6)(-10) = 21 - 5 - 60 = -44 \neq 0$.
So,the planes are not perpendicular.
Check for parallelism:
$\frac{A_1}{A_2} = \frac{7}{3}, \frac{B_1}{B_2} = \frac{5}{-1} = -5, \frac{C_1}{C_2} = \frac{6}{-10} = -0.6$.
Since $\frac{7}{3} \neq -5 \neq -0.6$,the planes are not parallel.
Calculate the angle $\theta$:
$\cos \theta = \left| \frac{-44}{\sqrt{7^2 + 5^2 + 6^2} \sqrt{3^2 + (-1)^2 + (-10)^2}} \right|$
$\cos \theta = \left| \frac{-44}{\sqrt{49 + 25 + 36} \sqrt{9 + 1 + 100}} \right| = \left| \frac{-44}{\sqrt{110} \sqrt{110}} \right| = \frac{44}{110} = \frac{2}{5}$.
Therefore,$\theta = \cos^{-1} \left( \frac{2}{5} \right)$.

(B) The direction ratios of the normal to the plane $L_1: a_1x + b_1y + c_1z + d_1 = 0$ are $(a_1, b_1, c_1)$ and for $L_2: a_2x + b_2y + c_2z + d_2 = 0$ are $(a_2, b_2, c_2)$.
$L_1 \parallel L_2$ if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
$L_1 \perp L_2$ if $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
The angle $\theta$ between $L_1$ and $L_2$ is given by $\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
The equations of the planes are $2x + y + 3z - 2 = 0$ and $x - 2y + 0z + 5 = 0$.
Here,$a_1 = 2, b_1 = 1, c_1 = 3$ and $a_2 = 1, b_2 = -2, c_2 = 0$.
Calculating the dot product of the normal vectors: $a_1a_2 + b_1b_2 + c_1c_2 = (2)(1) + (1)(-2) + (3)(0) = 2 - 2 + 0 = 0$.
Since the dot product is $0$,the given planes are perpendicular to each other.

(A) The direction ratios of the normals to the planes $L_1: a_1x + b_1y + c_1z + d_1 = 0$ and $L_2: a_2x + b_2y + c_2z + d_2 = 0$ are $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ respectively.
$L_1 \parallel L_2$ if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
$L_1 \perp L_2$ if $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
The angle $\theta$ between $L_1$ and $L_2$ is given by $\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
For the given planes $2x - 2y + 4z + 5 = 0$ and $3x - 3y + 6z - 1 = 0$:
$a_1 = 2, b_1 = -2, c_1 = 4$ and $a_2 = 3, b_2 = -3, c_2 = 6$.
Checking for parallelism:
$\frac{a_1}{a_2} = \frac{2}{3}, \frac{b_1}{b_2} = \frac{-2}{-3} = \frac{2}{3}, \frac{c_1}{c_2} = \frac{4}{6} = \frac{2}{3}$.
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{2}{3}$,the given planes are parallel to each other.

(A) The direction ratios of the normal to the plane $L_1: a_1x + b_1y + c_1z + d_1 = 0$ are $(a_1, b_1, c_1)$ and for $L_2: a_2x + b_2y + c_2z + d_2 = 0$ are $(a_2, b_2, c_2)$.
$L_1 \parallel L_2$ if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
$L_1 \perp L_2$ if $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
The angle $\theta$ between $L_1$ and $L_2$ is given by $\theta = \cos^{-1} \left( \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}} \right)$.
The equations of the planes are $2x - y + 3z - 1 = 0$ and $2x - y + 3z + 3 = 0$.
Here,$a_1 = 2, b_1 = -1, c_1 = 3$ and $a_2 = 2, b_2 = -1, c_2 = 3$.
Calculating the ratios: $\frac{a_1}{a_2} = \frac{2}{2} = 1$,$\frac{b_1}{b_2} = \frac{-1}{-1} = 1$,and $\frac{c_1}{c_2} = \frac{3}{3} = 1$.
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = 1$,the given planes are parallel to each other.

(D) The direction ratios of the normal to the planes $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0$ are $(A_1, B_1, C_1)$ and $(A_2, B_2, C_2)$ respectively.
The planes are parallel if $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$.
The planes are perpendicular if $A_1A_2 + B_1B_2 + C_1C_2 = 0$.
The angle $\theta$ between the planes is given by $\cos \theta = \left| \frac{A_1A_2 + B_1B_2 + C_1C_2}{\sqrt{A_1^2 + B_1^2 + C_1^2} \sqrt{A_2^2 + B_2^2 + C_2^2}} \right|$.
For the given planes $4x + 8y + z - 8 = 0$ and $0x + y + z - 4 = 0$:
$A_1 = 4, B_1 = 8, C_1 = 1$ and $A_2 = 0, B_2 = 1, C_2 = 1$.
Check for perpendicularity: $A_1A_2 + B_1B_2 + C_1C_2 = (4)(0) + (8)(1) + (1)(1) = 0 + 8 + 1 = 9 \neq 0$. Thus,they are not perpendicular.
Check for parallelism: $\frac{A_1}{A_2} = \frac{4}{0}$ (undefined),$\frac{B_1}{B_2} = 8$,$\frac{C_1}{C_2} = 1$. Since the ratios are not equal,they are not parallel.
Calculate the angle: $\cos \theta = \left| \frac{9}{\sqrt{4^2 + 8^2 + 1^2} \sqrt{0^2 + 1^2 + 1^2}} \right| = \left| \frac{9}{\sqrt{16 + 64 + 1} \sqrt{2}} \right| = \left| \frac{9}{9 \times \sqrt{2}} \right| = \frac{1}{\sqrt{2}}$.
Therefore,$\theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^{\circ}$ or $\frac{\pi}{4}$ radians.

(A) The distance $d$ of a point $P(x_1, y_1, z_1)$ from a plane $Ax + By + Cz - D = 0$ is given by the formula:
$d = \left| \frac{Ax_1 + By_1 + Cz_1 - D}{\sqrt{A^2 + B^2 + C^2}} \right|$
Given the point $(x_1, y_1, z_1) = (0, 0, 0)$ and the plane equation $3x - 4y + 12z - 3 = 0$,we identify $A = 3, B = -4, C = 12$,and $D = 3$.
Substituting these values into the formula:
$d = \left| \frac{3(0) - 4(0) + 12(0) - 3}{\sqrt{3^2 + (-4)^2 + 12^2}} \right|$
$d = \left| \frac{-3}{\sqrt{9 + 16 + 144}} \right|$
$d = \left| \frac{-3}{\sqrt{169}} \right|$
$d = \frac{3}{13}$

(A) The distance $d$ of a point $P(x_1, y_1, z_1)$ from a plane $Ax + By + Cz + D = 0$ is given by the formula:
$d = \left| \frac{Ax_1 + By_1 + Cz_1 + D}{\sqrt{A^2 + B^2 + C^2}} \right|$
Given the point $(x_1, y_1, z_1) = (3, -2, 1)$ and the plane equation $2x - y + 2z + 3 = 0$,we have $A = 2, B = -1, C = 2, D = 3$.
Substituting these values into the formula:
$d = \left| \frac{2(3) + (-1)(-2) + 2(1) + 3}{\sqrt{2^2 + (-1)^2 + 2^2}} \right|$
$d = \left| \frac{6 + 2 + 2 + 3}{\sqrt{4 + 1 + 4}} \right|$
$d = \left| \frac{13}{\sqrt{9}} \right|$
$d = \frac{13}{3}$

(A) The distance $d$ of a point $P(x_1, y_1, z_1)$ from a plane $Ax + By + Cz + D = 0$ is given by the formula:
$d = \left| \frac{Ax_1 + By_1 + Cz_1 + D}{\sqrt{A^2 + B^2 + C^2}} \right|$
Given point $P = (-6, 0, 0)$ and plane equation $2x - 3y + 6z - 2 = 0$.
Here,$A = 2, B = -3, C = 6, D = -2$ and $x_1 = -6, y_1 = 0, z_1 = 0$.
Substituting these values into the formula:
$d = \left| \frac{2(-6) - 3(0) + 6(0) - 2}{\sqrt{2^2 + (-3)^2 + 6^2}} \right|$
$d = \left| \frac{-12 - 0 + 0 - 2}{\sqrt{4 + 9 + 36}} \right|$
$d = \left| \frac{-14}{\sqrt{49}} \right|$
$d = \frac{14}{7} = 2$
Thus,the distance is $2$ units.

(A) The equation of a plane passing through the point $(x_1, y_1, z_1)$ is given by $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$.
Substituting the point $(1, -1, 2)$,we get $A(x - 1) + B(y + 1) + C(z - 2) = 0$ ... $(1)$.
Since this plane is perpendicular to the planes $2x + 3y - 2z = 5$ and $x + 2y - 3z = 8$,the normal vector of our plane $(A, B, C)$ must be perpendicular to the normals of the given planes $(2, 3, -2)$ and $(1, 2, -3)$.
Thus,the normal vector is the cross product of $(2, 3, -2)$ and $(1, 2, -3)$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -2 \\ 1 & 2 & -3 \end{vmatrix} = \hat{i}(-9 - (-4)) - \hat{j}(-6 - (-2)) + \hat{k}(4 - 3) = -5\hat{i} + 4\hat{j} + 1\hat{k}$.
So,$A = -5, B = 4, C = 1$.
Substituting these into $(1)$: $-5(x - 1) + 4(y + 1) + 1(z - 2) = 0$.
$-5x + 5 + 4y + 4 + z - 2 = 0$.
$-5x + 4y + z + 7 = 0$,which simplifies to $5x - 4y - z = 7$.

(A) First,find the equation of the plane passing through $A(3,-1,2)$,$B(5,2,4)$,and $C(-1,-1,6)$.
The vectors $\overline{AB} = (5-3)\hat{i} + (2-(-1))\hat{j} + (4-2)\hat{k} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ and $\overline{AC} = (-1-3)\hat{i} + (-1-(-1))\hat{j} + (6-2)\hat{k} = -4\hat{i} + 0\hat{j} + 4\hat{k}$.
The normal vector to the plane is $\vec{n} = \overline{AB} \times \overline{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 2 \\ -4 & 0 & 4 \end{vmatrix} = \hat{i}(12-0) - \hat{j}(8 - (-8)) + \hat{k}(0 - (-12)) = 12\hat{i} - 16\hat{j} + 12\hat{k}$.
Simplifying the normal vector by dividing by $4$,we get $\vec{n}' = 3\hat{i} - 4\hat{j} + 3\hat{k}$.
The equation of the plane is $3(x-3) - 4(y+1) + 3(z-2) = 0$,which simplifies to $3x - 4y + 3z - 19 = 0$.
The distance $d$ from point $P(6,5,9)$ to the plane $ax+by+cz+d=0$ is given by $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$.
Substituting the values: $d = \frac{|3(6) - 4(5) + 3(9) - 19|}{\sqrt{3^2 + (-4)^2 + 3^2}} = \frac{|18 - 20 + 27 - 19|}{\sqrt{9 + 16 + 9}} = \frac{|6|}{\sqrt{34}} = \frac{6}{\sqrt{34}} = \frac{6\sqrt{34}}{34} = \frac{3\sqrt{34}}{17}$.

(A) Any plane parallel to the plane $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=2$ is of the form $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=\lambda$.
Since the plane passes through the point $(a, b, c)$,its position vector is $\vec{r}=a\hat{i}+b\hat{j}+c\hat{k}$.
Substituting this into the equation,we get $(a\hat{i}+b\hat{j}+c\hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})=\lambda$,which implies $a+b+c=\lambda$.
Substituting $\lambda=a+b+c$ back into the equation,we get $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=a+b+c$.
In Cartesian form,substituting $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$,we obtain $x+y+z=a+b+c$.

(A) The equation of the plane passing through the point $(-1, 3, 2)$ is given by $a(x + 1) + b(y - 3) + c(z - 2) = 0$ $(1)$.
Where $a, b, c$ are the direction ratios of the normal to the plane.
Since the plane is perpendicular to $x + 2y + 3z = 5$,we have $a(1) + b(2) + c(3) = 0$,which implies $a + 2b + 3c = 0$ $(2)$.
Since the plane is also perpendicular to $3x + 3y + z = 0$,we have $a(3) + b(3) + c(1) = 0$,which implies $3a + 3b + c = 0$ $(3)$.
Solving equations $(2)$ and $(3)$ using cross-multiplication:
$\frac{a}{(2)(1) - (3)(3)} = \frac{b}{(3)(3) - (1)(1)} = \frac{c}{(1)(3) - (2)(3)}$
$\frac{a}{2 - 9} = \frac{b}{9 - 1} = \frac{c}{3 - 6}$
$\frac{a}{-7} = \frac{b}{8} = \frac{c}{-3} = k$
Thus,$a = -7k, b = 8k, c = -3k$.
Substituting these into $(1)$:
$-7k(x + 1) + 8k(y - 3) - 3k(z - 2) = 0$
$-7x - 7 + 8y - 24 - 3z + 6 = 0$
$-7x + 8y - 3z - 25 = 0$
Multiplying by $-1$,we get $7x - 8y + 3z + 25 = 0$.

(A) Let $P(x, y, z)$ be the point that is equidistant from points $A(1, 2, 3)$ and $B(3, 2, -1)$.
Accordingly,$PA = PB$.
$\Rightarrow PA^2 = PB^2$.
$\Rightarrow (x - 1)^2 + (y - 2)^2 + (z - 3)^2 = (x - 3)^2 + (y - 2)^2 + (z + 1)^2$.
Expanding both sides:
$x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9 = x^2 - 6x + 9 + y^2 - 4y + 4 + z^2 + 2z + 1$.
Simplifying the equation:
$-2x - 6z + 14 = -6x + 2z + 14$.
$-2x + 6x - 6z - 2z = 0$.
$4x - 8z = 0$.
Dividing by $4$,we get $x - 2z = 0$.
Thus,the required equation is $x - 2z = 0$.

(A) Let $P(x, y, z)$ be any point such that $PA = PB$.
Using the distance formula,we have:
$\sqrt{(x-3)^2 + (y-4)^2 + (z+5)^2} = \sqrt{(x+2)^2 + (y-1)^2 + (z-4)^2}$
Squaring both sides:
$(x-3)^2 + (y-4)^2 + (z+5)^2 = (x+2)^2 + (y-1)^2 + (z-4)^2$
Expanding the squares:
$(x^2 - 6x + 9) + (y^2 - 8y + 16) + (z^2 + 10z + 25) = (x^2 + 4x + 4) + (y^2 - 2y + 1) + (z^2 - 8z + 16)$
Simplifying the equation:
$-6x - 8y + 10z + 50 = 4x - 2y - 8z + 21$
Rearranging the terms:
$10x + 6y - 18z - 29 = 0$

(A) The coordinates of the origin $O$ are $(0, 0, 0)$ and the point $P$ are $(1, 2, -3).$
The direction ratios of the line segment $OP$ are $(1-0, 2-0, -3-0) = (1, 2, -3).$
Since the plane is perpendicular to $OP,$ the direction ratios of the normal to the plane are the same as the direction ratios of $OP,$ which are $a=1, b=2, c=-3.$
The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal direction ratios $(a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0.$
Substituting the point $P(1, 2, -3)$ and the direction ratios $(1, 2, -3)$ into the formula:
$1(x-1) + 2(y-2) - 3(z - (-3)) = 0$
$1(x-1) + 2(y-2) - 3(z+3) = 0$
$x - 1 + 2y - 4 - 3z - 9 = 0$
$x + 2y - 3z - 14 = 0.$

The equation of the plane having intercepts $a, b, c$ with $x, y, z$ axes respectively is given by:
$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ ..........$(1)$
The distance $p$ of the plane from the origin $(0, 0, 0)$ is given by the formula for the distance of a point from a plane $Ax + By + Cz + D = 0$,which is $p = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$.
Here,$A = \frac{1}{a}$,$B = \frac{1}{b}$,$C = \frac{1}{c}$,and $D = -1$.
Substituting these values:
$p = \frac{|\frac{1}{a}(0) + \frac{1}{b}(0) + \frac{1}{c}(0) - 1|}{\sqrt{(\frac{1}{a})^2 + (\frac{1}{b})^2 + (\frac{1}{c})^2}}$
$p = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$
Squaring both sides:
$p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}$
Taking the reciprocal:
$\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$

(C) The equations of the planes are:
$2x + 3y + 4z = 4$ $(1)$
$4x + 6y + 8z = 12$
Dividing the second equation by $2$,we get:
$2x + 3y + 4z = 6$ $(2)$
Since the coefficients of $x, y,$ and $z$ are the same in both equations,the planes are parallel.
The distance $D$ between two parallel planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$ is given by the formula:
$D = \left| \frac{d_2 - d_1}{\sqrt{a^2 + b^2 + c^2}} \right|$
Here,$a = 2, b = 3, c = 4, d_1 = 4,$ and $d_2 = 6$.
Substituting these values into the formula:
$D = \left| \frac{6 - 4}{\sqrt{2^2 + 3^2 + 4^2}} \right|$
$D = \left| \frac{2}{\sqrt{4 + 9 + 16}} \right|$
$D = \frac{2}{\sqrt{29}}$ units.
Thus,the correct answer is $C$.

(D) The equations of the planes are:
$2x - y + 4z = 5$ $(1)$
$5x - 2.5y + 10z = 6$ $(2)$
For two planes $a_1x + b_1y + c_1z = d_1$ and $a_2x + b_2y + c_2z = d_2$,they are parallel if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Calculating the ratios of the coefficients of $x, y,$ and $z$:
$\frac{a_1}{a_2} = \frac{2}{5} = 0.4$
$\frac{b_1}{b_2} = \frac{-1}{-2.5} = \frac{1}{2.5} = 0.4$
$\frac{c_1}{c_2} = \frac{4}{10} = 0.4$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = 0.4$,the normal vectors of the planes are proportional.
Also,check if the planes are coincident: $\frac{d_1}{d_2} = \frac{5}{6} \neq 0.4$. Since the ratios of the coefficients are equal but not equal to the ratio of the constants,the planes are parallel and distinct.
Therefore,the correct answer is $D$.

(A) Let the equation of the plane in intercept form be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
The coordinates of the points $A, B, C$ where the plane meets the axes are $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.
The centroid of $\Delta ABC$ is given by the formula $\left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3} \right)$.
Substituting the coordinates of $A, B, C$,the centroid is $\left( \frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3} \right) = \left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right)$.
We are given that the centroid is $(\alpha, \beta, \gamma)$.
Equating the coordinates,we get $\alpha = \frac{a}{3}$,$\beta = \frac{b}{3}$,and $\gamma = \frac{c}{3}$.
This implies $a = 3\alpha$,$b = 3\beta$,and $c = 3\gamma$.
Substituting these values into the intercept form equation $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,we get $\frac{x}{3\alpha} + \frac{y}{3\beta} + \frac{z}{3\gamma} = 1$.
Multiplying both sides by $3$,we obtain the required equation: $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3$.

(A) The plane bisects the line segment $AB$ at right angles,which means the plane passes through the midpoint of $AB$ and the vector $\vec{AB}$ acts as the normal vector $\vec{N}$ to the plane.
$1$. Midpoint $M$ of $AB$ is $\left(\frac{2+4}{2}, \frac{3+5}{2}, \frac{4+8}{2}\right) = (3, 4, 6)$.
$2$. The normal vector $\vec{N}$ is the vector $\vec{AB} = (4-2)\hat{i} + (5-3)\hat{j} + (8-4)\hat{k} = 2\hat{i} + 2\hat{j} + 4\hat{k}$.
$3$. The equation of a plane passing through point $\vec{a}$ with normal $\vec{N}$ is $(\vec{r} - \vec{a}) \cdot \vec{N} = 0$.
$4$. Substituting the values: $((x-3)\hat{i} + (y-4)\hat{j} + (z-6)\hat{k}) \cdot (2\hat{i} + 2\hat{j} + 4\hat{k}) = 0$.
$5$. Expanding the dot product: $2(x-3) + 2(y-4) + 4(z-6) = 0$.
$6$. Simplifying: $2x - 6 + 2y - 8 + 4z - 24 = 0 \Rightarrow 2x + 2y + 4z = 38$.
$7$. Dividing by $2$: $x + y + 2z = 19$.

(C) Let the normal vector to the plane be $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$.
Since the normal is equally inclined to the coordinate axes,the direction cosines are equal,i.e.,$\cos \alpha = \cos \beta = \cos \gamma$.
We know that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$,so $3 \cos^2 \alpha = 1$,which gives $\cos \alpha = \cos \beta = \cos \gamma = \frac{1}{\sqrt{3}}$.
Thus,the unit normal vector is $\hat{n} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}$.
The equation of a plane at a distance $p$ from the origin is given by $\vec{r} \cdot \hat{n} = p$.
Here,$p = 3\sqrt{3}$.
Substituting the values,we get $(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}) = 3\sqrt{3}$.
$\frac{x}{\sqrt{3}} + \frac{y}{\sqrt{3}} + \frac{z}{\sqrt{3}} = 3\sqrt{3}$.
Multiplying both sides by $\sqrt{3}$,we get $x + y + z = 3\sqrt{3} \times \sqrt{3} = 9$.
Therefore,the equation of the plane is $x + y + z = 9$.

(3X - 2Y + 6Z - 27 = 0) The line passes through $P(-2, -1, -3)$ and meets the plane at $Q(1, -3, 3)$ at a right angle.
Thus,the vector $\vec{PQ}$ is normal to the plane.
$\vec{PQ} = (1 - (-2))\hat{i} + (-3 - (-1))\hat{j} + (3 - (-3))\hat{k} = 3\hat{i} - 2\hat{j} + 6\hat{k}$.
The plane passes through the point $Q(1, -3, 3)$.
The equation of the plane passing through $(x_0, y_0, z_0)$ with normal $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.
Substituting the values: $3(x - 1) - 2(y + 3) + 6(z - 3) = 0$.
$3x - 3 - 2y - 6 + 6z - 18 = 0$.
$3x - 2y + 6z - 27 = 0$.

(N/A) The equation of a plane passing through three non-collinear points $(x_{1}, y_{1}, z_{1})$,$(x_{2}, y_{2}, z_{2})$,and $(x_{3}, y_{3}, z_{3})$ is given by the determinant equation:
$\begin{vmatrix} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{vmatrix} = 0$
Substituting the given points $(2, 1, 0)$,$(3, -2, -2)$,and $(3, 1, 7)$:
$\begin{vmatrix} x-2 & y-1 & z-0 \\ 3-2 & -2-1 & -2-0 \\ 3-2 & 1-1 & 7-0 \end{vmatrix} = 0$
Simplifying the matrix:
$\begin{vmatrix} x-2 & y-1 & z \\ 1 & -3 & -2 \\ 1 & 0 & 7 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$(x-2)(-21 - 0) - (y-1)(7 - (-2)) + z(0 - (-3)) = 0$
$(x-2)(-21) - (y-1)(9) + z(3) = 0$
$-21x + 42 - 9y + 9 + 3z = 0$
$-21x - 9y + 3z + 51 = 0$
Dividing the entire equation by $-3$:
$7x + 3y - z = 17$
Thus,the required equation of the plane is $7x + 3y - z = 17$.

The coordinates of the origin $O$ are $(0, 0, 0)$ and point $A$ are $(a, b, c)$.
The direction ratios of the line $OA$ are $(a-0, b-0, c-0) = (a, b, c)$.
The length of $OA$ is $\sqrt{a^2 + b^2 + c^2}$.
Therefore,the direction cosines of the line $OA$ are $\frac{a}{\sqrt{a^2 + b^2 + c^2}}, \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \frac{c}{\sqrt{a^2 + b^2 + c^2}}$.
The normal vector $\vec{n}$ to the plane is $\overrightarrow{OA} = a\hat{i} + b\hat{j} + c\hat{k}$.
The equation of a plane passing through a point $\vec{a} = (a, b, c)$ with normal vector $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$.
Substituting $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$,$\vec{a} = a\hat{i} + b\hat{j} + c\hat{k}$,and $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$:
$(x\hat{i} + y\hat{j} + z\hat{k} - (a\hat{i} + b\hat{j} + c\hat{k})) \cdot (a\hat{i} + b\hat{j} + c\hat{k}) = 0$
$(x-a)a + (y-b)b + (z-c)c = 0$
$ax - a^2 + by - b^2 + cz - c^2 = 0$
$ax + by + cz = a^2 + b^2 + c^2$.

(N/A) Let the two systems of rectangular axes be $S_1$ and $S_2$ with the same origin $O$.
The equation of the plane in the first system $S_1$ is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
The equation of the plane in the second system $S_2$ is given by $\frac{x}{a^{\prime}} + \frac{y}{b^{\prime}} + \frac{z}{c^{\prime}} = 1$.
The perpendicular distance $p$ from the origin $(0, 0, 0)$ to a plane $\frac{x}{A} + \frac{y}{B} + \frac{z}{C} = 1$ is given by $p = \frac{|-1|}{\sqrt{\frac{1}{A^2} + \frac{1}{B^2} + \frac{1}{C^2}}} = \frac{1}{\sqrt{\frac{1}{A^2} + \frac{1}{B^2} + \frac{1}{C^2}}}$.
Since the plane is the same,the perpendicular distance $p$ from the origin must be the same for both systems.
Therefore,$\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = \frac{1}{\sqrt{\frac{1}{a^{\prime 2}} + \frac{1}{b^{\prime 2}} + \frac{1}{c^{\prime 2}}}}$.
Squaring both sides,we get $\frac{1}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}} = \frac{1}{\frac{1}{a^{\prime 2}} + \frac{1}{b^{\prime 2}} + \frac{1}{c^{\prime 2}}}$.
Taking the reciprocal of both sides,we obtain $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{a^{\prime 2}} + \frac{1}{b^{\prime 2}} + \frac{1}{c^{\prime 2}}$.
Hence proved.

(A) The equation of the given plane is $ax + by = 0$ ... $(i)$.
The equation of the plane $z = 0$ is ... $(ii)$.
Any plane passing through the line of intersection of planes $(i)$ and $(ii)$ can be written as $ax + by + kz = 0$ ... $(iii)$.
The normal to plane $(i)$ is $\vec{n_1} = (a, b, 0)$ and the normal to plane $(iii)$ is $\vec{n_2} = (a, b, k)$.
The angle $\alpha$ between these two planes is given by $\cos \alpha = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
$\cos \alpha = \frac{|a^2 + b^2 + 0|}{\sqrt{a^2 + b^2} \sqrt{a^2 + b^2 + k^2}} = \frac{a^2 + b^2}{\sqrt{a^2 + b^2} \sqrt{a^2 + b^2 + k^2}} = \sqrt{\frac{a^2 + b^2}{a^2 + b^2 + k^2}}$.
Squaring both sides: $\cos^2 \alpha = \frac{a^2 + b^2}{a^2 + b^2 + k^2}$.
$a^2 + b^2 + k^2 = \frac{a^2 + b^2}{\cos^2 \alpha} = (a^2 + b^2) \sec^2 \alpha$.
$k^2 = (a^2 + b^2) \sec^2 \alpha - (a^2 + b^2) = (a^2 + b^2) (\sec^2 \alpha - 1) = (a^2 + b^2) \tan^2 \alpha$.
Thus,$k = \pm \sqrt{a^2 + b^2} \tan \alpha$.
Substituting $k$ into equation $(iii)$,we get $ax + by \pm (\sqrt{a^2 + b^2} \tan \alpha) z = 0$.

(D) The equation of a plane passing through the origin $(0, 0, 0)$ and containing the $y$-axis (which has direction vector $\hat{j}$) and passing through the point $P(1, 2, 3)$ can be found using the normal vector $\vec{n}$.
The plane contains the vector $\vec{v} = \hat{j} = (0, 1, 0)$ and the vector $\vec{OP} = \hat{i} + 2\hat{j} + 3\hat{k} = (1, 2, 3)$.
The normal vector $\vec{n}$ is given by the cross product:
$\vec{n} = \vec{v} \times \vec{OP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 0 \\ 1 & 2 & 3 \end{vmatrix}$
$\vec{n} = \hat{i}(3 - 0) - \hat{j}(0 - 0) + \hat{k}(0 - 1) = 3\hat{i} - \hat{k}$.
The equation of the plane passing through $(0, 0, 0)$ with normal $\vec{n} = (3, 0, -1)$ is:
$3(x - 0) + 0(y - 0) - 1(z - 0) = 0$
$3x - z = 0$.

(B) The normal vector $\overrightarrow{n}$ to the plane is perpendicular to both lines. Thus,$\overrightarrow{n}$ is the cross product of the direction vectors of the two lines:
$\overrightarrow{n} = (\hat{i} - 2\hat{j} + 2\hat{k}) \times (2\hat{i} + 3\hat{j} - \hat{k})$
$\overrightarrow{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 2 & 3 & -1 \end{vmatrix} = \hat{i}(2 - 6) - \hat{j}(-1 - 4) + \hat{k}(3 + 4) = -4\hat{i} + 5\hat{j} + 7\hat{k}$
The equation of the plane passing through $(3, 1, 1)$ with normal vector $\overrightarrow{n} = -4\hat{i} + 5\hat{j} + 7\hat{k}$ is:
$-4(x - 3) + 5(y - 1) + 7(z - 1) = 0$
$-4x + 12 + 5y - 5 + 7z - 7 = 0$
$-4x + 5y + 7z = 0$
Since the plane passes through $(\alpha, -3, 5)$,we substitute these coordinates into the plane equation:
$-4(\alpha) + 5(-3) + 7(5) = 0$
$-4\alpha - 15 + 35 = 0$
$-4\alpha + 20 = 0$
$4\alpha = 20$
$\alpha = 5$

(A) Let the points be $A(4, -2, 3)$ and $B(2, 4, -1)$.
The midpoint $M$ of the line segment $AB$ is given by:
$M = \left( \frac{4+2}{2}, \frac{-2+4}{2}, \frac{3-1}{2} \right) = (3, 1, 1)$.
The normal vector $\vec{n}$ to the plane is the vector $\vec{AB}$:
$\vec{n} = \vec{AB} = (2-4, 4-(-2), -1-3) = (-2, 6, -4)$.
We can simplify the normal vector by dividing by $-2$,so we take $\vec{n} = (1, -3, 2)$.
The equation of the plane passing through $M(3, 1, 1)$ with normal vector $\vec{n} = (1, -3, 2)$ is:
$1(x - 3) - 3(y - 1) + 2(z - 1) = 0$
$x - 3 - 3y + 3 + 2z - 2 = 0$
$x - 3y + 2z - 2 = 0$.
Now,we check which of the given points satisfies this equation:
For $(4, 0, -1)$: $4 - 3(0) + 2(-1) - 2 = 4 - 0 - 2 - 2 = 0$. This point satisfies the equation.
For $(4, 0, 1)$: $4 - 3(0) + 2(1) - 2 = 4 + 2 - 2 = 4 \neq 0$.
For $(0, 1, -1)$: $0 - 3(1) + 2(-1) - 2 = -3 - 2 - 2 = -7 \neq 0$.
For $(0, -1, 1)$: $0 - 3(-1) + 2(1) - 2 = 3 + 2 - 2 = 3 \neq 0$.
Thus,the plane passes through the point $(4, 0, -1)$.

(C) The midpoint $M$ of the line segment $PQ$ is given by $M = \left(\frac{4+2}{2}, \frac{-3+3}{2}, \frac{1-5}{2}\right) = (3, 0, -2)$.
Since the plane bisects the line segment $PQ$ at right angles,the normal vector to the plane is parallel to the vector $\vec{PQ} = (2-4, 3-(-3), -5-1) = (-2, 6, -6)$.
We can take the normal vector as $\vec{n} = (1, -3, 3)$ by dividing by $-2$.
The equation of the plane passing through $M(3, 0, -2)$ with normal vector $\vec{n} = (1, -3, 3)$ is:
$1(x-3) - 3(y-0) + 3(z+2) = 0$
$x - 3y + 3z - 3 + 6 = 0$
$x - 3y + 3z + 3 = 0$
Comparing this with $ax+by+cz+d=0$,we get $a=1, b=-3, c=3, d=3$.
The value of $(a^{2}+b^{2}+c^{2}+d^{2})$ is $1^{2} + (-3)^{2} + 3^{2} + 3^{2} = 1 + 9 + 9 + 9 = 28$.
Since $a, b, c, d$ are integers and the plane equation is unique up to a scalar multiple,the minimum value for integer coefficients is $28$.

(A) The equation of any plane parallel to $x - 2y + 2z - 3 = 0$ is of the form $x - 2y + 2z + \lambda = 0$.
The distance from the point $(1, 2, 3)$ to the plane $x - 2y + 2z + \lambda = 0$ is given by $\frac{|1 - 2(2) + 2(3) + \lambda|}{\sqrt{1^2 + (-2)^2 + 2^2}} = 1$.
Simplifying the expression: $\frac{|1 - 4 + 6 + \lambda|}{\sqrt{9}} = 1 \Rightarrow \frac{|\lambda + 3|}{3} = 1$.
This gives $|\lambda + 3| = 3$,so $\lambda + 3 = 3$ or $\lambda + 3 = -3$.
Thus,$\lambda = 0$ or $\lambda = -6$.
The two possible planes are $x - 2y + 2z = 0$ and $x - 2y + 2z - 6 = 0$.
Case $1$: $a=1, b=-2, c=2, d=0$. Then $(b - d) = -2 - 0 = -2$ and $(c - a) = 2 - 1 = 1$. So,$-2 = K(1) \Rightarrow K = -2$.
Case $2$: $a=1, b=-2, c=2, d=-6$. Then $(b - d) = -2 - (-6) = 4$ and $(c - a) = 2 - 1 = 1$. So,$4 = K(1) \Rightarrow K = 4$.
The positive value of $K$ is $4$.

(B) The equation of a plane passing through $(42, 0, 0)$,$(0, 42, 0)$,and $(0, 0, 42)$ in intercept form is $\frac{x}{42} + \frac{y}{42} + \frac{z}{42} = 1$,which simplifies to $x + y + z = 42$.
We can rewrite this as $(x-11) + (y-19) + (z-12) = 42 - 11 - 19 - 12 = 0$.
Let $a = x-11$,$b = y-19$,and $c = z-12$. Then $a + b + c = 0$.
The given expression is $3 + \frac{a}{b^2 c^2} + \frac{b}{a^2 c^2} + \frac{c}{a^2 b^2} - \frac{42}{14abc}$.
Since $a+b+c=0$,we have $x+y+z=42$. Substituting this into the last term,we get $\frac{42}{14abc} = \frac{3}{abc}$.
The expression becomes $3 + \frac{a^3 + b^3 + c^3}{a^2 b^2 c^2} - \frac{3}{abc} = 3 + \frac{a^3 + b^3 + c^3 - 3abc}{a^2 b^2 c^2}$.
Since $a+b+c=0$,the identity $a^3 + b^3 + c^3 = 3abc$ holds.
Therefore,the expression simplifies to $3 + \frac{3abc - 3abc}{a^2 b^2 c^2} = 3 + 0 = 3$.

(A) Let the point $P$ be $(x, y, z)$. The distances from the planes $x + y + z = 0$,$lx - nz = 0$,and $x - 2y + z = 0$ are given by $d_1 = \frac{|x + y + z|}{\sqrt{3}}$,$d_2 = \frac{|lx - nz|}{\sqrt{l^2 + n^2}}$,and $d_3 = \frac{|x - 2y + z|}{\sqrt{6}}$.
Given $d_1^2 + d_2^2 + d_3^2 = 9$,we have:
$\frac{(x + y + z)^2}{3} + \frac{(lx - nz)^2}{l^2 + n^2} + \frac{(x - 2y + z)^2}{6} = 9$.
Expanding the terms:
$\frac{x^2 + y^2 + z^2 + 2xy + 2yz + 2zx}{3} + \frac{l^2x^2 - 2lnxz + n^2z^2}{l^2 + n^2} + \frac{x^2 + 4y^2 + z^2 - 4xy - 4yz + 2zx}{6} = 9$.
Grouping the coefficients of $x^2, y^2, z^2, xy, yz, zx$:
$x^2(\frac{1}{3} + \frac{l^2}{l^2 + n^2} + \frac{1}{6}) + y^2(\frac{1}{3} + \frac{4}{6}) + z^2(\frac{1}{3} + \frac{n^2}{l^2 + n^2} + \frac{1}{6}) + xy(\frac{2}{3} - \frac{4}{6}) + yz(\frac{2}{3} - \frac{4}{6}) + zx(\frac{2}{3} - \frac{2ln}{l^2 + n^2} + \frac{2}{6}) = 9$.
Simplifying coefficients:
$x^2(\frac{1}{2} + \frac{l^2}{l^2 + n^2}) + y^2(1) + z^2(\frac{1}{2} + \frac{n^2}{l^2 + n^2}) + zx(1 - \frac{2ln}{l^2 + n^2}) = 9$.
Comparing this with $x^2 + y^2 + z^2 = 9$,the coefficients of $xy, yz, zx$ must be $0$ and the coefficients of $x^2, y^2, z^2$ must be $1$.
For the coefficient of $zx$ to be $0$,$1 - \frac{2ln}{l^2 + n^2} = 0 \implies l^2 + n^2 = 2ln \implies (l - n)^2 = 0 \implies l = n$.
Thus,$l - n = 0$.

(B) The equation of the plane is $2x - y + z = b$.
Let $P = (1, 3, a)$ and $Q = (-3, 5, 2)$. The midpoint $R$ of the line segment $PQ$ lies on the plane.
$R = \left( \frac{1 - 3}{2}, \frac{3 + 5}{2}, \frac{a + 2}{2} \right) = (-1, 4, \frac{a + 2}{2})$.
Since $R$ lies on the plane $2x - y + z = b$,we have:
$2(-1) - 4 + \frac{a + 2}{2} = b$
$-6 + \frac{a + 2}{2} = b \Rightarrow a + 2 = 2b + 12 \Rightarrow a = 2b + 10 \quad \dots(i)$
Also,the vector $\vec{PQ} = (-3 - 1, 5 - 3, 2 - a) = (-4, 2, 2 - a)$ is parallel to the normal vector of the plane $\vec{n} = (2, -1, 1)$.
Therefore,$\frac{-4}{2} = \frac{2}{-1} = \frac{2 - a}{1}$.
$-2 = -2 = 2 - a \Rightarrow a = 4$.
Substituting $a = 4$ into equation $(i)$:
$4 = 2b + 10 \Rightarrow 2b = -6 \Rightarrow b = -3$.
Thus,$|a + b| = |4 + (-3)| = |1| = 1$.

(C) The normal vector $\vec{n}$ of the required plane is perpendicular to the normal vectors of the given planes,$\vec{n_1} = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{n_2} = 2\hat{i} - 5\hat{j} - \hat{k}$.
Thus,$\vec{n} = \vec{n_1} \times \vec{n_2} = \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 2 & -5 & -1\end{array}\right|$.
Calculating the determinant:
$\vec{n} = \hat{i}(-1 - 10) - \hat{j}(-3 - (-4)) + \hat{k}(-15 - 2) = -11\hat{i} - \hat{j} - 17\hat{k}$.
We can take the normal vector as $\vec{n} = 11\hat{i} + \hat{j} + 17\hat{k}$ (multiplying by $-1$).
The equation of the plane passing through $(x_0, y_0, z_0) = (1, 2, -3)$ with normal vector $\langle a, b, c \rangle = \langle 11, 1, 17 \rangle$ is given by $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.
Substituting the values:
$11(x - 1) + 1(y - 2) + 17(z + 3) = 0$
$11x - 11 + y - 2 + 17z + 51 = 0$
$11x + y + 17z + 38 = 0$.

(B) The equation of plane $P_{1}$ is $3x + 15y + 21z = 9$. Dividing by $3$,we get $x + 5y + 7z = 3$.
The equation of plane $P_{2}$ is $x - 3y - z = 5$.
The equation of plane $P_{3}$ is $2x + 10y + 14z = 5$. Dividing by $2$,we get $x + 5y + 7z = \frac{5}{2}$.
Two planes $a_{1}x + b_{1}y + c_{1}z = d_{1}$ and $a_{2}x + b_{2}y + c_{2}z = d_{2}$ are parallel if their normal vectors are proportional,i.e.,$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$.
Comparing $P_{1}$ and $P_{3}$,the normal vectors are $(1, 5, 7)$ and $(1, 5, 7)$,which are identical. Since the constant terms $3$ and $\frac{5}{2}$ are different,the planes are parallel and distinct.
Therefore,$P_{1}$ and $P_{3}$ are parallel.

(C) The normal vector $\vec{n}$ to the plane is parallel to the line segment $\vec{AB}$.
$\vec{AB} = (-1 - (-2))\hat{i} + (-16 - (-21))\hat{j} + (33 - 29)\hat{k} = 1\hat{i} + 5\hat{j} + 4\hat{k}$.
The plane passes through $Q(4, -2, 2)$ and $P(\lambda, 2, 1)$. Thus,the vector $\vec{PQ}$ lies in the plane.
$\vec{PQ} = (4 - \lambda)\hat{i} + (-2 - 2)\hat{j} + (2 - 1)\hat{k} = (4 - \lambda)\hat{i} - 4\hat{j} + 1\hat{k}$.
Since the plane is perpendicular to the line $AB$,the normal vector $\vec{AB}$ is perpendicular to any vector in the plane,including $\vec{PQ}$.
Therefore,$\vec{AB} \cdot \vec{PQ} = 0$.
$(1\hat{i} + 5\hat{j} + 4\hat{k}) \cdot ((4 - \lambda)\hat{i} - 4\hat{j} + 1\hat{k}) = 0$.
$1(4 - \lambda) + 5(-4) + 4(1) = 0$.
$4 - \lambda - 20 + 4 = 0$.
$-\lambda - 12 = 0 \Rightarrow \lambda = -12$.
Now,calculate the expression $\left(\frac{\lambda}{11}\right)^{2} - \frac{4\lambda}{11} - 4$.
Substitute $\lambda = -12$: $\left(\frac{-12}{11}\right)^{2} - \frac{4(-12)}{11} - 4 = \frac{144}{121} + \frac{48}{11} - 4 = \frac{144 + 528 - 484}{121} = \frac{188}{121}$.

(C) The vectors lying on the plane are $\vec{AB} = (2-1)\hat{i} + (3-2)\hat{j} + (1-3)\hat{k} = \hat{i} + \hat{j} - 2\hat{k}$ and $\vec{AC} = (2-1)\hat{i} + (4-2)\hat{j} + (2-3)\hat{k} = \hat{i} + 2\hat{j} - \hat{k}$.
The normal vector $\vec{n}$ to the plane is given by $\vec{AB} \times \vec{AC} = \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \\ 1 & 2 & -1\end{array}\right| = \hat{i}(-1+4) - \hat{j}(-1+2) + \hat{k}(2-1) = 3\hat{i} - \hat{j} + \hat{k}$.
The vector $\vec{OP} = 2\hat{i} - \hat{j} + \hat{k}$.
Let $\theta$ be the angle between $\vec{OP}$ and the normal $\vec{n}$. Then $\cos \theta = \frac{|\vec{OP} \cdot \vec{n}|}{|\vec{OP}| |\vec{n}|} = \frac{|(2)(3) + (-1)(-1) + (1)(1)|}{\sqrt{2^2 + (-1)^2 + 1^2} \sqrt{3^2 + (-1)^2 + 1^2}} = \frac{|6 + 1 + 1|}{\sqrt{6} \sqrt{11}} = \frac{8}{\sqrt{66}}$.
Since $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{64}{66} = \frac{2}{66} = \frac{1}{33}$,we have $\sin \theta = \sqrt{\frac{1}{33}}$.
The length of the projection of $\vec{OP}$ on the plane is $|\vec{OP}| \sin \theta = \sqrt{6} \times \sqrt{\frac{1}{33}} = \sqrt{\frac{6}{33}} = \sqrt{\frac{2}{11}}$.

(C) The equation of the family of planes passing through the intersection of the given planes is given by $(x + y + 4z - 16) + \lambda(-x + y + z - 6) = 0$.
Since the plane $P$ passes through the point $(1, 2, 3)$,we substitute $x = 1, y = 2, z = 3$ into the equation:
$(1 + 2 + 4(3) - 16) + \lambda(-1 + 2 + 3 - 6) = 0$
$(1 + 2 + 12 - 16) + \lambda(-2) = 0$
$-1 - 2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$.
Substituting $\lambda = -\frac{1}{2}$ back into the family equation:
$(x + y + 4z - 16) - \frac{1}{2}(-x + y + z - 6) = 0$
$2(x + y + 4z - 16) - (-x + y + z - 6) = 0$
$2x + 2y + 8z - 32 + x - y - z + 6 = 0$
$3x + y + 7z - 26 = 0$.
Now,we check which point does not satisfy the equation $3x + y + 7z = 26$:
For $(3, 3, 2)$: $3(3) + 3 + 7(2) = 9 + 3 + 14 = 26$ (Lies on $P$).
For $(6, -6, 2)$: $3(6) - 6 + 7(2) = 18 - 6 + 14 = 26$ (Lies on $P$).
For $(4, 2, 2)$: $3(4) + 2 + 7(2) = 12 + 2 + 14 = 28 \neq 26$ (Does $NOT$ lie on $P$).
For $(-8, 8, 6)$: $3(-8) + 8 + 7(6) = -24 + 8 + 42 = 26$ (Lies on $P$).

(A) Since $R(3,5,\gamma)$ lies on the plane $2x-y+z+3=0$,we have:
$2(3) - 5 + \gamma + 3 = 0$
$6 - 5 + \gamma + 3 = 0$
$4 + \gamma = 0 \Rightarrow \gamma = -4$.
Thus,$R$ is $(3,5,-4)$.
The normal vector to the plane is $\vec{n} = (2, -1, 1)$. The line $QS$ passes through $Q(1,3,4)$ and is parallel to $\vec{n}$.
The equation of line $QS$ is $\frac{x-1}{2} = \frac{y-3}{-1} = \frac{z-4}{1} = \lambda$.
Any point on this line is $F(2\lambda+1, -\lambda+3, \lambda+4)$.
Since $F$ is the foot of the perpendicular from $Q$ to the plane,it lies on the plane:
$2(2\lambda+1) - (-\lambda+3) + (\lambda+4) + 3 = 0$
$4\lambda + 2 + \lambda - 3 + \lambda + 4 + 3 = 0$
$6\lambda + 6 = 0 \Rightarrow \lambda = -1$.
Substituting $\lambda = -1$ into $F$,we get $F(-1, 4, 3)$.
Since $F$ is the midpoint of $QS$,let $S = (x_s, y_s, z_s)$:
$\frac{x_s+1}{2} = -1 \Rightarrow x_s = -3$
$\frac{y_s+3}{2} = 4 \Rightarrow y_s = 5$
$\frac{z_s+4}{2} = 3 \Rightarrow z_s = 2$.
So,$S = (-3, 5, 2)$.
The square of the length $SR$ is:
$SR^2 = (3 - (-3))^2 + (5 - 5)^2 + (-4 - 2)^2$
$SR^2 = (6)^2 + (0)^2 + (-6)^2 = 36 + 0 + 36 = 72$.

(B) The equations of the planes are $P_{1}: x-2y-2z+1=0$ and $P_{2}: 2x-3y-6z+1=0$.
The equation of the angle bisector is given by $\left|\frac{x-2y-2z+1}{\sqrt{1^2+(-2)^2+(-2)^2}}\right| = \left|\frac{2x-3y-6z+1}{\sqrt{2^2+(-3)^2+(-6)^2}}\right|$.
This simplifies to $\frac{x-2y-2z+1}{3} = \pm \frac{2x-3y-6z+1}{7}$.
To determine the acute angle bisector,we check the sign of $a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = (1)(2) + (-2)(-3) + (-2)(-6) = 2 + 6 + 12 = 20$.
Since $20 > 0$,the negative sign gives the acute angle bisector.
Thus,$\frac{x-2y-2z+1}{3} = -\frac{2x-3y-6z+1}{7}$.
$7(x-2y-2z+1) = -3(2x-3y-6z+1)$.
$7x-14y-14z+7 = -6x+9y+18z-3$.
$13x-23y-32z+10 = 0$.
Testing the point $\left(-2, 0, -\frac{1}{2}\right)$: $13(-2) - 23(0) - 32(-\frac{1}{2}) + 10 = -26 - 0 + 16 + 10 = 0$.
Therefore,the point $\left(-2, 0, -\frac{1}{2}\right)$ lies on the plane $P$.

(B) The normal vector $\vec{n}$ of the required plane is perpendicular to the normals of the given planes $\vec{n}_1 = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{n}_2 = \hat{i} - \hat{j} - \hat{k}$.
Thus,$\vec{n} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & -1 & -1 \end{vmatrix} = \hat{i}(-1 - 1) - \hat{j}(-2 + 1) + \hat{k}(-2 - 1) = -2\hat{i} + \hat{j} - 3\hat{k}$.
The equation of the plane passing through $(-1, 0, -2)$ with normal vector $\vec{n} = -2\hat{i} + \hat{j} - 3\hat{k}$ is:
$-2(x + 1) + 1(y - 0) - 3(z + 2) = 0$
$-2x - 2 + y - 3z - 6 = 0$
$-2x + y - 3z - 8 = 0$
Multiplying by $-1$,we get $2x - y + 3z + 8 = 0$.
Comparing this with $ax + by + cz + 8 = 0$,we have $a = 2$,$b = -1$,and $c = 3$.
Therefore,$a + b + c = 2 - 1 + 3 = 4$.

(C) The plane $P$ is the perpendicular bisector plane of the line segment joining $A(-4, 2, 1)$ and $B(2, -2, 3)$.
The normal vector $\vec{n}_1$ of plane $P$ is the vector $\vec{AB} = (2 - (-4))\hat{i} + (-2 - 2)\hat{j} + (3 - 1)\hat{k} = 6\hat{i} - 4\hat{j} + 2\hat{k}$.
We can take the normal vector as $\vec{n}_1 = 3\hat{i} - 2\hat{j} + \hat{k}$.
The midpoint of $AB$ is $M = \left(\frac{-4+2}{2}, \frac{2-2}{2}, \frac{1+3}{2}\right) = (-1, 0, 2)$.
The equation of plane $P$ is $3(x + 1) - 2(y - 0) + 1(z - 2) = 0$,which simplifies to $3x - 2y + z + 1 = 0$.
The second plane is $P': 2x + y + 3z - 1 = 0$,with normal vector $\vec{n}_2 = 2\hat{i} + \hat{j} + 3\hat{k}$.
The angle $\theta$ between the two planes is given by $\cos \theta = \left| \frac{\vec{n}_1 \cdot \vec{n}_2}{|\vec{n}_1| |\vec{n}_2|} \right|$.
$\vec{n}_1 \cdot \vec{n}_2 = (3)(2) + (-2)(1) + (1)(3) = 6 - 2 + 3 = 7$.
$|\vec{n}_1| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$.
$|\vec{n}_2| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$.
$\cos \theta = \left| \frac{7}{\sqrt{14} \cdot \sqrt{14}} \right| = \frac{7}{14} = \frac{1}{2}$.
Therefore,$\theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$.

(D) The normal vector $\vec{n}$ of the required plane is perpendicular to the normal vectors of the given planes,$\vec{n_1} = 2\hat{i} + \hat{j} - 5\hat{k}$ and $\vec{n_2} = 3\hat{i} + 5\hat{j} - 7\hat{k}$.
Thus,$\vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -5 \\ 3 & 5 & -7 \end{vmatrix} = \hat{i}(-7 + 25) - \hat{j}(-14 + 15) + \hat{k}(10 - 3) = 18\hat{i} - \hat{j} + 7\hat{k}$.
The equation of the plane is $18x - y + 7z = d$.
Since the plane passes through $(2, 3, -5)$,we have $18(2) - (3) + 7(-5) = d$,which gives $36 - 3 - 35 = d$,so $d = -2$.
The equation of the plane is $18x - y + 7z = -2$,or $-18x + y - 7z = 2$.
Comparing this with $ax + by + cz = d$,we get $a = -18, b = 1, c = -7, d = 2$.
Since $\text{gcd}(|-18|, |1|, |-7|, 2) = 1$ and $d > 0$,these values are correct.
Finally,$a + 7b + c + 20d = -18 + 7(1) + (-7) + 20(2) = -18 + 7 - 7 + 40 = 22$.

(C) The formula for the mirror image $(a, b, c)$ of a point $(x_1, y_1, z_1)$ in the plane $Ax + By + Cz + D = 0$ is given by $\frac{a - x_1}{A} = \frac{b - y_1}{B} = \frac{c - z_1}{C} = \frac{-2(Ax_1 + By_1 + Cz_1 + D)}{A^2 + B^2 + C^2}$.
Given the point $(2, 4, 7)$ and the plane $3x - y + 4z - 2 = 0$,we have $A=3, B=-1, C=4, D=-2$.
Substituting these values into the formula:
$\frac{a - 2}{3} = \frac{b - 4}{-1} = \frac{c - 7}{4} = \frac{-2(3(2) - 1(4) + 4(7) - 2)}{3^2 + (-1)^2 + 4^2}$
$\frac{a - 2}{3} = \frac{b - 4}{-1} = \frac{c - 7}{4} = \frac{-2(6 - 4 + 28 - 2)}{9 + 1 + 16} = \frac{-2(28)}{26} = \frac{-56}{26} = \frac{-28}{13}$.
Now,solving for $a, b, c$:
$a = 2 + 3(\frac{-28}{13}) = 2 - \frac{84}{13} = \frac{26 - 84}{13} = \frac{-58}{13}$.
$b = 4 - 1(\frac{-28}{13}) = 4 + \frac{28}{13} = \frac{52 + 28}{13} = \frac{80}{13}$.
$c = 7 + 4(\frac{-28}{13}) = 7 - \frac{112}{13} = \frac{91 - 112}{13} = \frac{-21}{13}$.
Finally,calculating $2a + b + 2c$:
$2(\frac{-58}{13}) + \frac{80}{13} + 2(\frac{-21}{13}) = \frac{-116 + 80 - 42}{13} = \frac{-78}{13} = -6$.