Plane Questions in English

(D) The equation of the plane is $3x + 4y + 12z - 52 = 0$.
The length of the perpendicular $p$ from a point $(x_1, y_1, z_1)$ to the plane $Ax + By + Cz + D = 0$ is given by the formula $p = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.
Here,the point is the origin $(0, 0, 0)$,so $x_1 = 0, y_1 = 0, z_1 = 0$.
Substituting the values,we get $p = \frac{|3(0) + 4(0) + 12(0) - 52|}{\sqrt{3^2 + 4^2 + 12^2}}$.
$p = \frac{|-52|}{\sqrt{9 + 16 + 144}} = \frac{52}{\sqrt{169}} = \frac{52}{13} = 4$.
Since the value $4$ is not among the options $A, B, C$,the correct option is $D$.

(D) The coordinates of point $P$ are $(2, 6, 3)$ and the origin $O$ is $(0, 0, 0)$.
The vector $\vec{OP}$ acts as the normal vector to the plane,so $\vec{n} = \vec{OP} = 2\hat{i} + 6\hat{j} + 3\hat{k}$.
The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$.
Substituting the values $a=2, b=6, c=3$ and $(x_1, y_1, z_1) = (2, 6, 3)$:
$2(x - 2) + 6(y - 6) + 3(z - 3) = 0$
Expanding the equation:
$2x - 4 + 6y - 36 + 3z - 9 = 0$
$2x + 6y + 3z - 49 = 0$
$2x + 6y + 3z = 49$.
Thus,the correct option is $D$.

(C) To find the intercepts of the plane on the coordinate axes,we convert the equation into the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Given equation: $5x - 3y + 6z = 60$.
Divide the entire equation by $60$:
$\frac{5x}{60} - \frac{3y}{60} + \frac{6z}{60} = \frac{60}{60}$
Simplifying the fractions:
$\frac{x}{12} - \frac{y}{20} + \frac{z}{10} = 1$
Rewriting in the standard intercept form:
$\frac{x}{12} + \frac{y}{-20} + \frac{z}{10} = 1$
Comparing this with $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,we get the intercepts as $a = 12$,$b = -20$,and $c = 10$.
Thus,the intercepts are $(12, -20, 10)$.

(C) The general equation of a plane is given by $ax + by + cz + d = 0$,where $(a, b, c)$ are the direction ratios of the normal to the plane.
If a plane is parallel to the $x$-axis,its normal must be perpendicular to the $x$-axis.
The direction ratios of the $x$-axis are $(1, 0, 0)$.
Since the normal $(a, b, c)$ is perpendicular to the $x$-axis $(1, 0, 0)$,their dot product must be zero:
$a(1) + b(0) + c(0) = 0 \implies a = 0$.
Substituting $a = 0$ into the general equation,we get $0x + by + cz + d = 0$,which simplifies to $by + cz + d = 0$.
Therefore,the correct option is $C$.

(A) The equation of a plane passing through $A(-1, 3, 0)$ is given by $A(x + 1) + B(y - 3) + C(z - 0) = 0$ ... $(i)$.
Since the plane passes through $B(2, 2, 1)$ and $C(1, 1, 3)$,we have:
$A(2 + 1) + B(2 - 3) + C(1 - 0) = 0 \implies 3A - B + C = 0$ ... $(ii)$.
$A(1 + 1) + B(1 - 3) + C(3 - 0) = 0 \implies 2A - 2B + 3C = 0$ ... $(iii)$.
Solving $(ii)$ and $(iii)$ using cross-multiplication:
$\frac{A}{(-1)(3) - (1)(-2)} = \frac{B}{(1)(2) - (3)(3)} = \frac{C}{(3)(-2) - (-1)(2)}$
$\frac{A}{-3 + 2} = \frac{B}{2 - 9} = \frac{C}{-6 + 2} \implies \frac{A}{-1} = \frac{B}{-7} = \frac{C}{-4}$.
Thus,the direction ratios are $(1, 7, 4)$.
Substituting into $(i)$: $1(x + 1) + 7(y - 3) + 4(z) = 0 \implies x + 7y + 4z - 20 = 0$.
The distance from the point $D(5, 7, 8)$ to the plane $x + 7y + 4z - 20 = 0$ is:
$d = \frac{|1(5) + 7(7) + 4(8) - 20|}{\sqrt{1^2 + 7^2 + 4^2}} = \frac{|5 + 49 + 32 - 20|}{\sqrt{1 + 49 + 16}} = \frac{66}{\sqrt{66}} = \sqrt{66}$.

(B) The given equation is $x^2 - 5x + 6 = 0$.
Factoring the quadratic equation,we get $(x - 2)(x - 3) = 0$.
This implies $x = 2$ or $x = 3$.
In a three-dimensional $xyz$ coordinate system,the equation $x = k$ (where $k$ is a constant) represents a plane parallel to the $yz$-plane.
Therefore,$x = 2$ and $x = 3$ represent two distinct planes parallel to the $yz$-plane.
Thus,the equation represents a pair of planes.

(A) The equations $|x| = p$,$|y| = p$,and $|z| = p$ represent planes parallel to the coordinate planes.
Specifically,$|x| = p$ represents the two planes $x = p$ and $x = -p$.
Similarly,$|y| = p$ represents $y = p$ and $y = -p$,and $|z| = p$ represents $z = p$ and $z = -p$.
These six planes define the boundaries of a cube centered at the origin $(0, 0, 0)$ with side length $2p$ and faces parallel to the coordinate planes.
Therefore,the correct option is $A$.

(A) The given equation of the plane is $by + cz + d = 0$.
This can be written as $0 \cdot x + b \cdot y + c \cdot z + d = 0$.
The normal vector to this plane is $\vec{n_1} = (0, b, c)$.
The $YOZ$ plane (or $yz$-plane) has the equation $x = 0$,which can be written as $1 \cdot x + 0 \cdot y + 0 \cdot z = 0$.
The normal vector to the $YOZ$ plane is $\vec{n_2} = (1, 0, 0)$.
Two planes are perpendicular if the dot product of their normal vectors is zero.
$\vec{n_1} \cdot \vec{n_2} = (0 \times 1) + (b \times 0) + (c \times 0) = 0 + 0 + 0 = 0$.
Since the dot product is $0$,the plane $by + cz + d = 0$ is perpendicular to the $YOZ$ plane.

(A) The equation of a plane parallel to $ax + by + cz + d = 0$ is given by $ax + by + cz + k = 0$.
Given the plane $x + 2y + 5z = 0$,any plane parallel to it will have the form $x + 2y + 5z + k = 0$.
Since this plane passes through the point $(1, 2, 3)$,we substitute these coordinates into the equation:
$(1) + 2(2) + 5(3) + k = 0$
$1 + 4 + 15 + k = 0$
$20 + k = 0$
$k = -20$
Substituting $k = -20$ back into the equation,we get $x + 2y + 5z - 20 = 0$,or $x + 2y + 5z = 20$.
Comparing this with the given options,the equation $(x - 1) + 2(y - 2) + 5(z - 3) = 0$ simplifies to $x - 1 + 2y - 4 + 5z - 15 = 0$,which is $x + 2y + 5z - 20 = 0$. Thus,option $A$ is correct.

(B) The equation of any plane parallel to $5x - 6y + 7z = 3$ is of the form $5x - 6y + 7z = k$.
Since this plane passes through the point $(2, 3, 4)$,we substitute these coordinates into the equation:
$5(2) - 6(3) + 7(4) = k$
$10 - 18 + 28 = k$
$20 = k$
Substituting the value of $k$ back into the equation,we get $5x - 6y + 7z = 20$,which can be written as $5x - 6y + 7z - 20 = 0$.

(A) The distance of a plane $ax + by + cz + d = 0$ from the origin $(0, 0, 0)$ is given by the formula:
Distance $= \frac{|d|}{\sqrt{a^2 + b^2 + c^2}}$
Here,the equation of the plane is $6x - 3y + 2z - 14 = 0$.
Comparing this with the standard form,we have $a = 6$,$b = -3$,$c = 2$,and $d = -14$.
Substituting these values into the formula:
Distance $= \frac{|-14|}{\sqrt{6^2 + (-3)^2 + 2^2}}$
$= \frac{14}{\sqrt{36 + 9 + 4}}$
$= \frac{14}{\sqrt{49}}$
$= \frac{14}{7} = 2$.
Thus,the distance of the plane from the origin is $2$ units.

(A) For two planes $P_1: ax + by + cz + d = 0$ and $P_2: a'x + b'y + c'z + d' = 0$,the origin $(0, 0, 0)$ lies in the acute angle if the signs of the expressions $P_1(0,0,0)$ and $P_2(0,0,0)$ are the same as the signs of $aa' + bb' + cc'$ and the product of the constant terms $dd' > 0$.
Specifically,if $aa' + bb' + cc' < 0$,the origin lies in the acute angle if $d$ and $d'$ have the same sign,which implies $dd' > 0$.

(C) The given equations of the planes are $2x + y + 2z - 8 = 0$ and $4x + 2y + 4z + 5 = 0$.
To find the distance between parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$,we first make the coefficients of $x, y, z$ identical.
Multiply the first equation by $2$: $4x + 2y + 4z - 16 = 0$.
Now,the planes are $4x + 2y + 4z - 16 = 0$ and $4x + 2y + 4z + 5 = 0$.
The distance $d$ is given by the formula $d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$.
Here,$A = 4, B = 2, C = 4, D_1 = -16, D_2 = 5$.
$d = \frac{|-16 - 5|}{\sqrt{4^2 + 2^2 + 4^2}} = \frac{|-21|}{\sqrt{16 + 4 + 16}} = \frac{21}{\sqrt{36}} = \frac{21}{6} = \frac{7}{2}$.

(A) The angle $\theta$ between two planes $a_1x + b_1y + c_1z = d_1$ and $a_2x + b_2y + c_2z = d_2$ is given by the formula:
$\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$
Here,the normal vectors are $\vec{n_1} = (1, 2, 2)$ and $\vec{n_2} = (-5, 3, 4)$.
Substituting the values:
$\cos \theta = \frac{|(1)(-5) + (2)(3) + (2)(4)|}{\sqrt{1^2 + 2^2 + 2^2} \sqrt{(-5)^2 + 3^2 + 4^2}}$
$\cos \theta = \frac{|-5 + 6 + 8|}{\sqrt{1 + 4 + 4} \sqrt{25 + 9 + 16}}$
$\cos \theta = \frac{9}{\sqrt{9} \sqrt{50}}$
$\cos \theta = \frac{9}{3 \times 5\sqrt{2}} = \frac{3}{5\sqrt{2}}$
Rationalizing the denominator:
$\cos \theta = \frac{3 \times \sqrt{2}}{5 \times 2} = \frac{3\sqrt{2}}{10}$
Therefore,$\theta = \cos^{-1} \left( \frac{3\sqrt{2}}{10} \right)$.

(B) The distance of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.
Given the plane $3x + 4y - 12z + 13 = 0$,the distance of point $P(1, 1, k)$ is $d_1 = \frac{|3(1) + 4(1) - 12(k) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|20 - 12k|}{\sqrt{9 + 16 + 144}} = \frac{|20 - 12k|}{13}$.
The distance of point $Q(-3, 0, 1)$ is $d_2 = \frac{|3(-3) + 4(0) - 12(1) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|-9 + 0 - 12 + 13|}{13} = \frac{|-8|}{13} = \frac{8}{13}$.
Since the points are equidistant,$d_1 = d_2$,so $\frac{|20 - 12k|}{13} = \frac{8}{13}$.
This implies $|20 - 12k| = 8$,which gives two cases:
Case $1$: $20 - 12k = 8 \Rightarrow 12k = 12 \Rightarrow k = 1$.
Case $2$: $20 - 12k = -8 \Rightarrow 12k = 28 \Rightarrow k = \frac{28}{12} = \frac{7}{3}$.
Since $k=1$ is provided in the options,the correct answer is $1$.

(C) The origin $O$ is $(0, 0, 0)$ and point $P$ is $(1, 2, -3)$.
Since the plane is perpendicular to $OP$,the vector $\vec{OP}$ acts as the normal vector $\vec{n}$ to the plane.
$\vec{n} = \vec{OP} = (1 - 0)\hat{i} + (2 - 0)\hat{j} + (-3 - 0)\hat{k} = 1\hat{i} + 2\hat{j} - 3\hat{k}$.
The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$.
Substituting the values $a=1, b=2, c=-3$ and $(x_1, y_1, z_1) = (1, 2, -3)$:
$1(x - 1) + 2(y - 2) - 3(z + 3) = 0$
$x - 1 + 2y - 4 - 3z - 9 = 0$
$x + 2y - 3z - 14 = 0$.

(B) The direction ratios of the line passing through the points $(x_1, y_1, z_1) = (1, 2, 0)$ and $(x_2, y_2, z_2) = (4, 13, 5)$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.
Substituting the values,we get $(4 - 1, 13 - 2, 5 - 0) = (3, 11, 5)$.
Since the line is perpendicular to the plane,the direction ratios of the line are the same as the direction ratios of the normal to the plane.
The equation of a plane with normal vector $(a, b, c)$ is $ax + by + cz + d = 0$.
Therefore,the coefficients of $x, y$ and $z$ in the equation of the plane are the direction ratios of the normal,which are $3, 11, 5$.

(D) The distance of the point $(1, 1, 1)$ from the origin $(0, 0, 0)$ is given by $\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.
The distance of the point $(1, 1, 1)$ from the plane $x + y + z + k = 0$ is given by $\frac{|1 + 1 + 1 + k|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{|k + 3|}{\sqrt{3}}$.
According to the problem,the distance from the origin is half the distance from the plane:
$\sqrt{3} = \frac{1}{2} \times \frac{|k + 3|}{\sqrt{3}}$.
Multiplying both sides by $2\sqrt{3}$,we get:
$2 \times 3 = |k + 3|$
$|k + 3| = 6$.
This implies $k + 3 = 6$ or $k + 3 = -6$.
Solving these,we get $k = 3$ or $k = -9$.

(B) Let the intercepts of the plane on the coordinate axes be $a, b$,and $c$. Thus,the coordinates of the points are $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.
The centroid of the triangle $ABC$ is given by $\left( \frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3} \right) = \left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right)$.
Given that the centroid is $(1, 2, 4)$,we have:
$\frac{a}{3} = 1 \implies a = 3$
$\frac{b}{3} = 2 \implies b = 6$
$\frac{c}{3} = 4 \implies c = 12$
The intercept form of the equation of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Substituting the values of $a, b$,and $c$,we get $\frac{x}{3} + \frac{y}{6} + \frac{z}{12} = 1$.
Multiplying the entire equation by $12$,we get $4x + 2y + z = 12$.

(A) The intercept form of the equation of a plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the intercepts on the $x, y, z$ axes respectively.
Given intercepts are $a = 8, b = 4, c = 4$.
Substituting these values,the equation of the plane is $\frac{x}{8} + \frac{y}{4} + \frac{z}{4} = 1$.
Multiplying by $8$,we get $x + 2y + 2z = 8$,or $x + 2y + 2z - 8 = 0$.
The length of the perpendicular from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.
Here,$A = 1, B = 2, C = 2, D = -8$.
Thus,$d = \frac{|-8|}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{8}{\sqrt{1 + 4 + 4}} = \frac{8}{\sqrt{9}} = \frac{8}{3}$.

(C) Let the points be $A(-1, 2, 3)$ and $B(3, -5, 6)$.
The normal vector $\vec{n}$ to the plane is the vector $\vec{AB} = (3 - (-1), -5 - 2, 6 - 3) = (4, -7, 3)$.
The midpoint $M$ of the line segment $AB$ is $\left( \frac{-1+3}{2}, \frac{2-5}{2}, \frac{3+6}{2} \right) = \left( 1, -\frac{3}{2}, \frac{9}{2} \right)$.
The equation of the plane passing through $(x_0, y_0, z_0)$ with normal vector $(a, b, c)$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.
Substituting the values: $4(x - 1) - 7(y + \frac{3}{2}) + 3(z - \frac{9}{2}) = 0$.
$4x - 4 - 7y - \frac{21}{2} + 3z - \frac{27}{2} = 0$.
$4x - 7y + 3z - 4 - \frac{48}{2} = 0$.
$4x - 7y + 3z - 4 - 24 = 0$.
$4x - 7y + 3z = 28$.

(A) The line passes through the points $P(4, -1, 2)$ and $Q(-3, 2, 3)$.
The direction ratios of this line are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1) = (-3 - 4, 2 - (-1), 3 - 2) = (-7, 3, -1)$.
Since the line is perpendicular to the plane,the direction ratios of the normal to the plane are the same as the direction ratios of the line,which are $(-7, 3, -1)$.
Alternatively,we can use the normal vector $\vec{n} = (7, -3, 1)$ by multiplying by $-1$.
The plane passes through the point $R(-10, 5, 4)$.
The equation of a plane passing through $(x_0, y_0, z_0)$ with normal vector $(a, b, c)$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.
Substituting the values,we get $7(x - (-10)) - 3(y - 5) - 1(z - 4) = 0$.
$7(x + 10) - 3(y - 5) - (z - 4) = 0$.
$7x + 70 - 3y + 15 - z + 4 = 0$.
$7x - 3y - z + 89 = 0$.

(A) The plane that bisects the line segment joining two points must pass through the midpoint of that segment.
Let the points be $A(2, 3, 4)$ and $B(6, 7, 8)$.
The midpoint $M$ of $AB$ is given by the formula $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)$.
Substituting the coordinates,we get $M = \left( \frac{2 + 6}{2}, \frac{3 + 7}{2}, \frac{4 + 8}{2} \right) = (4, 5, 6)$.
Now,we check which of the given options is satisfied by the point $(4, 5, 6)$:
For option $A$: $x + y + z - 15 = 0 \implies 4 + 5 + 6 - 15 = 15 - 15 = 0$. This is satisfied.
Thus,the equation of the plane is $x + y + z - 15 = 0$.

(C) The direction ratios of the line are $(3, 0, 4)$. Since the plane is perpendicular to this line,these direction ratios are the normal vector $(A, B, C)$ of the plane.
The equation of a plane passing through $(x_1, y_1, z_1)$ with normal $(A, B, C)$ is $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$.
Substituting the given values: $3(x - 1) + 0(y - 1) + 4(z - 1) = 0$.
Simplifying,we get $3x - 3 + 4z - 4 = 0$,which is $3x + 4z - 7 = 0$.
The perpendicular distance $d$ from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.
Here,$A = 3, B = 0, C = 4, D = -7$.
$d = \frac{|-7|}{\sqrt{3^2 + 0^2 + 4^2}} = \frac{7}{\sqrt{9 + 16}} = \frac{7}{\sqrt{25}} = \frac{7}{5}$.

(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 $(2, -1, -3)$,we get $A(x - 2) + B(y + 1) + C(z + 3) = 0$.
Since the plane is parallel to the lines with direction ratios $(3, 2, -4)$ and $(2, -3, 2)$,the normal vector $(A, B, C)$ must be perpendicular to both direction vectors.
Thus,$3A + 2B - 4C = 0$ and $2A - 3B + 2C = 0$.
Using the cross product to find the normal vector $(A, B, C)$:
$A = (2)(2) - (-3)(-4) = 4 - 12 = -8$
$B = -((3)(2) - (2)(-4)) = -(6 + 8) = -14$
$C = (3)(-3) - (2)(2) = -9 - 4 = -13$
So,the normal vector is $(-8, -14, -13)$.
Substituting these into the plane equation: $-8(x - 2) - 14(y + 1) - 13(z + 3) = 0$.
Multiplying by $-1$,we get $8(x - 2) + 14(y + 1) + 13(z + 3) = 0$.
Expanding this: $8x - 16 + 14y + 14 + 13z + 39 = 0$.
Simplifying,we get $8x + 14y + 13z + 37 = 0$.

(B) Let the coordinates of point $P$ be $(x, y, z)$.
Given $PA^2 - PB^2 = k$.
Using the distance formula,$PA^2 = (x - 2)^2 + (y - 3)^2 + (z - 4)^2$ and $PB^2 = (x + 2)^2 + (y - 5)^2 + (z + 4)^2$.
Substituting these into the equation:
$[(x - 2)^2 + (y - 3)^2 + (z - 4)^2] - [(x + 2)^2 + (y - 5)^2 + (z + 4)^2] = k$
Expanding the squares:
$(x^2 - 4x + 4 + y^2 - 6y + 9 + z^2 - 8z + 16) - (x^2 + 4x + 4 + y^2 - 10y + 25 + z^2 + 8z + 16) = k$
Simplifying the expression:
$(x^2 + y^2 + z^2 - 4x - 6y - 8z + 29) - (x^2 + y^2 + z^2 + 4x - 10y + 8z + 45) = k$
$-8x + 4y - 16z - 16 = k$
This is a linear equation in $x, y, z$ of the form $ax + by + cz + d = 0$,which represents a plane.

(A) Let the normal vector to the required plane be $\vec{n} = l\hat{i} + m\hat{j} + n\hat{k}$.
Since the plane is perpendicular to $x + 2y + 2z = 5$ and $3x + 3y + 2z = 8$,the normal vector $\vec{n}$ is perpendicular to the normal vectors $\vec{n}_1 = (1, 2, 2)$ and $\vec{n}_2 = (3, 3, 2)$.
Thus,$\vec{n} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 3 & 3 & 2 \end{vmatrix} = \hat{i}(4 - 6) - \hat{j}(2 - 6) + \hat{k}(3 - 6) = -2\hat{i} + 4\hat{j} - 3\hat{k}$.
The equation of the plane passing through $(1, -3, -2)$ with normal vector $(-2, 4, -3)$ is:
$-2(x - 1) + 4(y + 3) - 3(z + 2) = 0$
$-2x + 2 + 4y + 12 - 3z - 6 = 0$
$-2x + 4y - 3z + 8 = 0$
Multiplying by $-1$,we get $2x - 4y + 3z - 8 = 0$.

(A) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the intercepts on the $x, y, z$ axes respectively.
The distance of this plane from the origin $(0, 0, 0)$ is given by $p = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$.
Squaring both sides,we get $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$.
The points of intersection with the axes are $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.
Planes drawn through these points parallel to the coordinate planes intersect at the point $P(x, y, z) = (a, b, c)$.
Substituting $a=x, b=y, c=z$ into the distance equation,the locus of the point of intersection is $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{1}{p^2}$.

(D) The line passes through the origin $(0, 0, 0)$ and is equally inclined to the axes,so its direction ratios are $(1, 1, 1)$.
The point $P$ is $(a, a, a)$. The vector $\vec{OP}$ is the normal to the plane,so the normal vector is $\vec{n} = a\hat{i} + a\hat{j} + a\hat{k}$,which is parallel to $\hat{i} + \hat{j} + \hat{k}$.
The equation of the plane passing through $(a, a, a)$ with normal vector $(1, 1, 1)$ is:
$1(x - a) + 1(y - a) + 1(z - a) = 0$
$x + y + z = 3a$
To find the intercepts,we write the equation in the intercept form $\frac{x}{X} + \frac{y}{Y} + \frac{z}{Z} = 1$:
$\frac{x}{3a} + \frac{y}{3a} + \frac{z}{3a} = 1$
The intercepts on the axes are $X = 3a$,$Y = 3a$,and $Z = 3a$.
The sum of the reciprocals of the intercepts is:
$\frac{1}{3a} + \frac{1}{3a} + \frac{1}{3a} = \frac{3}{3a} = \frac{1}{a}$.

(B) The equation of a plane passing through the intersection of two planes $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$.
Here,the planes are $x + 2y + 3z - 4 = 0$ and $4x + 3y + 2z + 1 = 0$.
So,the equation of the required plane is $(x + 2y + 3z - 4) + \lambda(4x + 3y + 2z + 1) = 0$.
Since this plane passes through the origin $(0, 0, 0)$,we substitute $x=0, y=0, z=0$ into the equation:
$(0 + 0 + 0 - 4) + \lambda(0 + 0 + 0 + 1) = 0$
$-4 + \lambda = 0 \implies \lambda = 4$.
Substituting $\lambda = 4$ back into the equation:
$(x + 2y + 3z - 4) + 4(4x + 3y + 2z + 1) = 0$
$x + 2y + 3z - 4 + 16x + 12y + 8z + 4 = 0$
$17x + 14y + 11z = 0$.

(B) The equation of a plane passing through $(1, 0, 0)$ and $(0, 1, 0)$ can be written in intercept form as $\frac{x}{1} + \frac{y}{1} + \frac{z}{c} = 1$,where $c$ is the $z$-intercept.
This simplifies to $x + y + \frac{z}{c} = 1$.
The direction ratios of the normal to this plane are $(1, 1, \frac{1}{c})$.
The given plane is $x + y = 3$,and its normal vector has direction ratios $(1, 1, 0)$.
The angle $\theta$ between two planes with normals $\vec{n_1} = (a_1, b_1, c_1)$ and $\vec{n_2} = (a_2, b_2, c_2)$ is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Given $\theta = \frac{\pi}{4}$,we have $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.
Substituting the values: $\frac{1}{\sqrt{2}} = \frac{|1(1) + 1(1) + \frac{1}{c}(0)|}{\sqrt{1^2 + 1^2 + (\frac{1}{c})^2} \sqrt{1^2 + 1^2 + 0^2}}$.
$\frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2 + \frac{1}{c^2}} \cdot \sqrt{2}}$.
$\frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2} \sqrt{2 + \frac{1}{c^2}}}$.
$1 = \frac{2}{\sqrt{2 + \frac{1}{c^2}}}$.
$\sqrt{2 + \frac{1}{c^2}} = 2$.
Squaring both sides: $2 + \frac{1}{c^2} = 4$.
$\frac{1}{c^2} = 2 \Rightarrow c^2 = \frac{1}{2} \Rightarrow c = \frac{1}{\sqrt{2}}$.
The direction ratios are $(1, 1, \frac{1}{c}) = (1, 1, \sqrt{2})$.

(D) Let the equation of the plane in the first system of rectangular axes be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
The perpendicular distance $p$ from the origin $(0, 0, 0)$ to this plane is given by $p = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$,which implies $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$.
Similarly,for the second system of rectangular axes,the equation of the same plane is $\frac{x}{a'} + \frac{y}{b'} + \frac{z}{c'} = 1$.
The perpendicular distance $p$ from the origin to this plane is the same,so $p = \frac{|-1|}{\sqrt{\frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2}}}$,which implies $\frac{1}{p^2} = \frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2}$.
Equating the two expressions for $\frac{1}{p^2}$,we get $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2}$.
Therefore,$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} - \frac{1}{a'^2} - \frac{1}{b'^2} - \frac{1}{c'^2} = 0$.

(B) The length of the perpendicular from a point $(x_1, y_1, z_1)$ to the plane $Ax + By + Cz + D = 0$ is given by $P = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.
For the plane $3x - 6y + 2z + 11 = 0$,the denominator is $\sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.
For point $(2, 3, 4)$,${P_1} = \frac{|3(2) - 6(3) + 2(4) + 11|}{7} = \frac{|6 - 18 + 8 + 11|}{7} = \frac{|7|}{7} = 1$.
For point $(1, 1, 4)$,${P_2} = \frac{|3(1) - 6(1) + 2(4) + 11|}{7} = \frac{|3 - 6 + 8 + 11|}{7} = \frac{|16|}{7} = \frac{16}{7}$.
The quadratic equation with roots ${P_1}$ and ${P_2}$ is given by $(P - P_1)(P - P_2) = 0$,which is $P^2 - (P_1 + P_2)P + P_1P_2 = 0$.
Sum of roots: $P_1 + P_2 = 1 + \frac{16}{7} = \frac{23}{7}$.
Product of roots: $P_1P_2 = 1 \times \frac{16}{7} = \frac{16}{7}$.
Substituting these into the equation: $P^2 - \frac{23}{7}P + \frac{16}{7} = 0$,which simplifies to $7P^2 - 23P + 16 = 0$.

(A) The angle between two faces of a tetrahedron is the angle between their normal vectors $n_1$ and $n_2$.
For face $OAB$,the normal vector $n_1$ is given by the cross product of vectors $\vec{OA}$ and $\vec{OB}$:
$n_1 = \vec{OA} \times \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(3-2) + \hat{k}(1-4) = 5\hat{i} - \hat{j} - 3\hat{k}$.
For face $ABC$,the normal vector $n_2$ is given by the cross product of vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AB} = (2-1)\hat{i} + (1-2)\hat{j} + (3-1)\hat{k} = \hat{i} - \hat{j} + 2\hat{k}$.
$\vec{AC} = (-1-1)\hat{i} + (1-2)\hat{j} + (2-1)\hat{k} = -2\hat{i} - \hat{j} + \hat{k}$.
$n_2 = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ -2 & -1 & 1 \end{vmatrix} = \hat{i}(-1+2) - \hat{j}(1+4) + \hat{k}(-1-2) = \hat{i} - 5\hat{j} - 3\hat{k}$.
The angle $\theta$ between the faces is the angle between the normals $n_1$ and $n_2$:
$\cos \theta = \frac{|n_1 \cdot n_2|}{|n_1| |n_2|} = \frac{|(5)(1) + (-1)(-5) + (-3)(-3)|}{\sqrt{5^2 + (-1)^2 + (-3)^2} \sqrt{1^2 + (-5)^2 + (-3)^2}} = \frac{|5 + 5 + 9|}{\sqrt{25+1+9} \sqrt{1+25+9}} = \frac{19}{\sqrt{35} \sqrt{35}} = \frac{19}{35}$.
Thus,$\theta = \cos^{-1}\left(\frac{19}{35}\right)$.

(C) The normal vector $\vec{n_1}$ of plane $P_1$ is $\langle 2, -1, 1 \rangle$ and the normal vector $\vec{n_2}$ of plane $P_2$ is $\langle 1, 2, -1 \rangle$.
Since the required plane is perpendicular to both $P_1$ and $P_2$,its normal vector $\vec{n}$ must be parallel to $\vec{n_1} \times \vec{n_2}$.
$\vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} = \hat{i}(1 - 2) - \hat{j}(-2 - 1) + \hat{k}(4 + 1) = -\hat{i} + 3\hat{j} + 5\hat{k}$.
Thus,the normal vector is $\langle -1, 3, 5 \rangle$.
The equation of the plane passing through $(-1, 3, 2)$ with normal $\langle -1, 3, 5 \rangle$ is given by:
$-1(x - (-1)) + 3(y - 3) + 5(z - 2) = 0$
$-1(x + 1) + 3(y - 3) + 5(z - 2) = 0$
$-x - 1 + 3y - 9 + 5z - 10 = 0$
$-x + 3y + 5z - 20 = 0$
Multiplying by $-1$,we get $x - 3y - 5z + 20 = 0$.

(D) The equations of the planes are $P_1 : 2x - y + z - 2 = 0$ and $P_2 : x + 2y - z - 3 = 0$.
To find the angle bisector,we use the formula $\frac{a_1x + b_1y + c_1z + d_1}{\sqrt{a_1^2 + b_1^2 + c_1^2}} = \pm \frac{a_2x + b_2y + c_2z + d_2}{\sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Substituting the values: $\frac{2x - y + z - 2}{\sqrt{4 + 1 + 1}} = \pm \frac{x + 2y - z - 3}{\sqrt{1 + 4 + 1}}$.
Since $\sqrt{6} = \sqrt{6}$,we have $2x - y + z - 2 = \pm (x + 2y - z - 3)$.
Case $1$ (Positive sign): $2x - y + z - 2 = x + 2y - z - 3 \Rightarrow x - 3y + 2z + 1 = 0$.
Case $2$ (Negative sign): $2x - y + z - 2 = -x - 2y + z + 3 \Rightarrow 3x + y - 5 = 0$.
To check which plane contains the origin $(0,0,0)$,we substitute $(0,0,0)$ into the equations:
For $x - 3y + 2z + 1 = 0$,$0 - 0 + 0 + 1 = 1 \neq 0$.
For $3x + y - 5 = 0$,$0 + 0 - 5 = -5 \neq 0$.
Since both do not contain the origin,we look for the bisector of the angle containing the origin. The sign of $d_1$ and $d_2$ should be the same. Here $d_1 = -2$ and $d_2 = -3$. Since they have the same sign,the positive sign gives the bisector of the angle containing the origin. Thus,the negative sign gives the bisector not containing the origin: $3x + y - 5 = 0$.

(D) The vector equation of a plane passing through a point with position vector $\vec{a}$ and perpendicular to a normal vector $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$,which can be written as $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
Here,$\vec{a} = 2\hat{i} + 3\hat{j} - 4\hat{k}$ and $\vec{n} = 2\hat{i} - \hat{j} + 2\hat{k}$.
Substituting these values into the equation:
$\vec{r} \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = (2\hat{i} + 3\hat{j} - 4\hat{k}) \cdot (2\hat{i} - \hat{j} + 2\hat{k})$
$\vec{r} \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = (2)(2) + (3)(-1) + (-4)(2)$
$\vec{r} \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = 4 - 3 - 8 = -7$.
To convert this to Cartesian form,let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.
$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = -7$
$2x - y + 2z = -7$.
Thus,the Cartesian equation of the plane is $2x - y + 2z = -7$.

(C) The angle $\theta$ between two planes $\vec{r} \cdot \vec{n_1} = d_1$ and $\vec{r} \cdot \vec{n_2} = d_2$ is given by $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
Here,$\vec{n_1} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{n_2} = \hat{i} + \hat{j} + 2\hat{k}$.
Calculating the dot product: $\vec{n_1} \cdot \vec{n_2} = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3$.
Calculating the magnitudes: $|\vec{n_1}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ and $|\vec{n_2}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.
Thus,$\cos \theta = \frac{|3|}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$.
Therefore,$\theta = \cos^{-1}(1/2) = \pi/3$.

(A) The normal vectors to the planes $2x - y + z = 6$ and $x + y + 2z = 3$ are $\vec{n_1} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{n_2} = \hat{i} + \hat{j} + 2\hat{k}$ respectively.
The angle $\theta$ between the two planes is given by the formula $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
Calculating the dot product: $\vec{n_1} \cdot \vec{n_2} = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3$.
Calculating the magnitudes: $|\vec{n_1}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ and $|\vec{n_2}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.
Substituting these values into the formula: $\cos \theta = \frac{3}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$.
Therefore,$\theta = \cos^{-1}(1/2) = \pi / 3$.

(C) The normal vectors to the given planes $2x - y + z = 6$ and $x + y + 2z = 3$ are $\vec{n_1} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{n_2} = \hat{i} + \hat{j} + 2\hat{k}$ respectively.
The angle $\theta$ between the two planes is given by the formula $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
Calculate the dot product: $\vec{n_1} \cdot \vec{n_2} = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3$.
Calculate the magnitudes: $|\vec{n_1}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ and $|\vec{n_2}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.
Substitute the values into the formula: $\cos \theta = \frac{|3|}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = 60^o$.

(A) The distance $d$ of a plane $Ax + By + Cz + D = 0$ from the origin $(0, 0, 0)$ is given by the formula:
$d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$
Here,$A = 2$,$B = -3$,$C = 6$,and $D = 14$.
Substituting these values into the formula:
$d = \frac{|14|}{\sqrt{2^2 + (-3)^2 + 6^2}}$
$d = \frac{14}{\sqrt{4 + 9 + 36}}$
$d = \frac{14}{\sqrt{49}}$
$d = \frac{14}{7} = 2$
Thus,the distance is $2$ units.

(A) The equation of the plane is $x + 2y - 3z + 4 = 0$.
Comparing this with the general form $Ax + By + Cz + D = 0$,we get the normal vector $\vec{n} = (1, 2, -3)$.
The magnitude of the normal vector is $|\vec{n}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
The direction cosines $(l, m, n)$ are given by $\frac{A}{|\vec{n}|}, \frac{B}{|\vec{n}|}, \frac{C}{|\vec{n}|}$.
Thus,the direction cosines are $\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{-3}{\sqrt{14}}$.
Alternatively,multiplying the equation by $-1$,we get $-x - 2y + 3z - 4 = 0$,which gives the direction cosines as $-\frac{1}{\sqrt{14}}, -\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$.
Comparing this with the given options,option $A$ is correct.

(B) The given line is parallel to the vector $\vec{n} = \hat{i} - \hat{j} + 2\hat{k}$.
Since the plane is perpendicular to this line,the vector $\vec{n} = \hat{i} - \hat{j} + 2\hat{k}$ acts as the normal vector to the plane.
The plane passes through the point $A(2, 3, 1)$,which has the position vector $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$.
The equation of a plane passing through a point $\vec{a}$ and having a normal vector $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$.
Substituting the values,we get $(\vec{r} - (2\hat{i} + 3\hat{j} + \hat{k})) \cdot (\hat{i} - \hat{j} + 2\hat{k}) = 0$.
$\vec{r} \cdot (\hat{i} - \hat{j} + 2\hat{k}) = (2\hat{i} + 3\hat{j} + \hat{k}) \cdot (\hat{i} - \hat{j} + 2\hat{k})$.
$\vec{r} \cdot (\hat{i} - \hat{j} + 2\hat{k}) = (2)(1) + (3)(-1) + (1)(2) = 2 - 3 + 2 = 1$.
Thus,the equation of the plane is $\vec{r} \cdot (\hat{i} - \hat{j} + 2\hat{k}) = 1$.

(A) The distance $D$ between two parallel planes $ax + by + cz + d_1 = 0$ and $ax + by + cz + d_2 = 0$ is given by the formula:
$D = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$
Given the planes $ax + by + cz + d = 0$ and $ax + by + cz + d' = 0$,we substitute $d_1 = d$ and $d_2 = d'$.
Therefore,the distance is $\frac{|d - d'|}{\sqrt{a^2 + b^2 + c^2}}$.

(C) The given equations of the planes are $2x + y + 2z - 8 = 0$ and $4x + 2y + 4z + 5 = 0$.
To make the coefficients of $x, y, z$ identical,multiply the first equation by $2$:
$4x + 2y + 4z - 16 = 0$
Now,the planes are $4x + 2y + 4z - 16 = 0$ and $4x + 2y + 4z + 5 = 0$.
The distance $d$ between two parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$ is given by $d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$.
Here,$A = 4, B = 2, C = 4, D_1 = -16, D_2 = 5$.
$d = \frac{|-16 - 5|}{\sqrt{4^2 + 2^2 + 4^2}} = \frac{|-21|}{\sqrt{16 + 4 + 16}} = \frac{21}{\sqrt{36}} = \frac{21}{6} = \frac{7}{2}$.

(A) The equation of a plane passing through a point $\vec{a} = (2, 3, -1)$ and perpendicular to a normal vector $\vec{n} = 3\hat{i} - 4\hat{j} + 7\hat{k}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$.
Substituting the values,we get $(x - 2)(3) + (y - 3)(-4) + (z + 1)(7) = 0$.
$3x - 6 - 4y + 12 + 7z + 7 = 0$.
$3x - 4y + 7z + 13 = 0$.
The distance $d$ of the plane $Ax + By + Cz + D = 0$ from the origin $(0, 0, 0)$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.
Here,$A = 3, B = -4, C = 7, D = 13$.
$d = \frac{|13|}{\sqrt{3^2 + (-4)^2 + 7^2}} = \frac{13}{\sqrt{9 + 16 + 49}} = \frac{13}{\sqrt{74}}$.

(A) The vector equation of a plane passing through a point with position vector $\vec{a}$ and normal to vector $\vec{n}$ is given by:
$(\vec{r} - \vec{a}) \cdot \vec{n} = 0$ or $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \dots (i)$
Given that the plane passes through the point $P(3, -3, 1)$,we have $\vec{a} = 3\hat{i} - 3\hat{j} + \hat{k}$.
The plane is normal to the line joining $A(3, 4, -1)$ and $B(2, -1, 5)$. Thus,the normal vector $\vec{n}$ is the vector $\vec{AB}$:
$\vec{n} = \vec{AB} = (2 - 3)\hat{i} + (-1 - 4)\hat{j} + (5 - (-1))\hat{k} = -\hat{i} - 5\hat{j} + 6\hat{k}$.
Substituting $\vec{a}$ and $\vec{n}$ into equation $(i)$:
$\vec{r} \cdot (-\hat{i} - 5\hat{j} + 6\hat{k}) = (3\hat{i} - 3\hat{j} + \hat{k}) \cdot (-\hat{i} - 5\hat{j} + 6\hat{k})$
$\vec{r} \cdot (-\hat{i} - 5\hat{j} + 6\hat{k}) = (3)(-1) + (-3)(-5) + (1)(6)$
$\vec{r} \cdot (-\hat{i} - 5\hat{j} + 6\hat{k}) = -3 + 15 + 6 = 18$.
To find the Cartesian equation,substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$:
$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (-\hat{i} - 5\hat{j} + 6\hat{k}) = 18$
$-x - 5y + 6z = 18$
$x + 5y - 6z + 18 = 0$.

(A) Let the equation of the plane be $\frac{X}{a} + \frac{Y}{b} + \frac{Z}{c} = 1$.
Since the plane is at a unit distance from the origin $(0, 0, 0)$,we have $\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 1$,which implies $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = 1$.
The coordinates of the points are $P(a, 0, 0)$,$Q(0, b, 0)$,and $R(0, 0, c)$.
The centroid $(x, y, z)$ of $\Delta PQR$ is given by $x = \frac{a+0+0}{3} = \frac{a}{3}$,$y = \frac{0+b+0}{3} = \frac{b}{3}$,and $z = \frac{0+0+c}{3} = \frac{c}{3}$.
Thus,$a = 3x$,$b = 3y$,and $c = 3z$.
Substituting these into the plane equation condition: $\frac{1}{(3x)^2} + \frac{1}{(3y)^2} + \frac{1}{(3z)^2} = 1$.
$\frac{1}{9x^2} + \frac{1}{9y^2} + \frac{1}{9z^2} = 1$.
Multiplying by $9$,we get $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 9$.
Therefore,$k = 9$.

(B) The equation of a plane parallel to $x + y + z = 0$ is of the form $x + y + z + \lambda = 0$,where $\lambda$ is a constant.
Since the plane passes through the point $(\alpha, \beta, \gamma)$,we substitute these coordinates into the equation:
$\alpha + \beta + \gamma + \lambda = 0$
Solving for $\lambda$,we get $\lambda = -(\alpha + \beta + \gamma)$.
Substituting the value of $\lambda$ back into the equation of the plane,we get:
$x + y + z - (\alpha + \beta + \gamma) = 0$
Rearranging the terms,we get $x + y + z = \alpha + \beta + \gamma$.

(C) The general equation of a plane passing through $A(2, 2, -1)$ is given by:
$a(x - 2) + b(y - 2) + c(z + 1) = 0 \dots(i)$
Since the plane passes through $B(3, 4, 2)$ and $C(7, 0, 6)$,we substitute these points into $(i)$:
For $B(3, 4, 2)$: $a(3 - 2) + b(4 - 2) + c(2 + 1) = 0 \implies a + 2b + 3c = 0 \dots(ii)$
For $C(7, 0, 6)$: $a(7 - 2) + b(0 - 2) + c(6 + 1) = 0 \implies 5a - 2b + 7c = 0 \dots(iii)$
Solving $(ii)$ and $(iii)$ using cross-multiplication:
$\frac{a}{(2)(7) - (3)(-2)} = \frac{b}{(3)(5) - (1)(7)} = \frac{c}{(1)(-2) - (2)(5)}$
$\frac{a}{14 + 6} = \frac{b}{15 - 7} = \frac{c}{-2 - 10}$
$\frac{a}{20} = \frac{b}{8} = \frac{c}{-12} \implies \frac{a}{5} = \frac{b}{2} = \frac{c}{-3} = k$
Thus,$a = 5k, b = 2k, c = -3k$. Substituting these into $(i)$:
$5k(x - 2) + 2k(y - 2) - 3k(z + 1) = 0$
$5(x - 2) + 2(y - 2) - 3(z + 1) = 0$
$5x - 10 + 2y - 4 - 3z - 3 = 0$
$5x + 2y - 3z = 17$.