Plane Questions in English

(A) Let the plane be $ax + by + cz = d$. The normal vector to the plane is the vector from the origin $(0, 0, 0)$ to the foot of the perpendicular $(2, -3, 6)$.
Thus,the normal vector $\vec{n} = (2, -3, 6)$.
The equation of the plane is $2x - 3y + 6z = d$.
Since the point $(2, -3, 6)$ lies on the plane,we substitute these coordinates into the equation:
$2(2) - 3(-3) + 6(6) = d$
$4 + 9 + 36 = d$
$d = 49$.
Therefore,the equation of the plane is $2x - 3y + 6z = 49$,which can be written as $2x - 3y + 6z - 49 = 0$.

(B) Let the direction ratios of the normal to the required plane be $\langle a, b, c \rangle$.
Since the plane passes through $(4,4,0)$,its equation is $a(x-4) + b(y-4) + c(z-0) = 0$.
This plane is perpendicular to the planes $2x+y+2z+3=0$ and $3x+3y+2z-8=0$.
Thus,the normal vector $\vec{n} = \langle a, b, c \rangle$ is perpendicular to the normals of the given planes,$\vec{n_1} = \langle 2, 1, 2 \rangle$ and $\vec{n_2} = \langle 3, 3, 2 \rangle$.
Therefore,$\vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 3 & 3 & 2 \end{vmatrix} = \hat{i}(2-6) - \hat{j}(4-6) + \hat{k}(6-3) = -4\hat{i} + 2\hat{j} + 3\hat{k}$.
So,the direction ratios are $\langle -4, 2, 3 \rangle$.
The equation of the plane is $-4(x-4) + 2(y-4) + 3(z-0) = 0$.
$-4x + 16 + 2y - 8 + 3z = 0$.
$-4x + 2y + 3z + 8 = 0$,which simplifies to $4x - 2y - 3z = 8$.

(D) The given vector equation is $\vec{r} = (1+\lambda-\mu) \hat{i} + (2-\lambda) \hat{j} + (3-2 \lambda+2 \mu) \hat{k}$.
Rearranging the terms,we get:
$\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} - 2\hat{k}) + \mu(-\hat{i} + 2\hat{k})$.
This is in the form $\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$,where $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{b} = \hat{i} - \hat{j} - 2\hat{k}$,and $\vec{c} = -\hat{i} + 2\hat{k}$.
The normal vector $\vec{n}$ to the plane is given by $\vec{n} = \vec{b} \times \vec{c}$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -2 \\ -1 & 0 & 2 \end{vmatrix} = \hat{i}(-2 - 0) - \hat{j}(2 - 2) + \hat{k}(0 - 1) = -2\hat{i} - \hat{k}$.
The equation of the plane is $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$,which implies $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
$\vec{a} \cdot \vec{n} = (\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (-2\hat{i} - \hat{k}) = (1)(-2) + (2)(0) + (3)(-1) = -2 - 3 = -5$.
So,$\vec{r} \cdot (-2\hat{i} - \hat{k}) = -5$,or $\vec{r} \cdot (2\hat{i} + \hat{k}) = 5$.
Substituting $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$,we get $(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} + \hat{k}) = 5$,which simplifies to $2x + z = 5$.

(C) The plane passes through the point $(x_1, y_1, z_1) = (3, -2, -1)$ and is parallel to vectors $\vec{b} = (1, -2, 4)$ and $\vec{c} = (3, 2, -5)$.
The Cartesian equation of a plane passing through $(x_1, y_1, z_1)$ and parallel to vectors $\vec{b} = (a_1, b_1, c_1)$ and $\vec{c} = (a_2, b_2, c_2)$ is given by the determinant equation:
$\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$
Substituting the given values:
$\begin{vmatrix} x - 3 & y + 2 & z + 1 \\ 1 & -2 & 4 \\ 3 & 2 & -5 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$(x - 3)((-2)(-5) - (4)(2)) - (y + 2)((1)(-5) - (4)(3)) + (z + 1)((1)(2) - (-2)(3)) = 0$
$(x - 3)(10 - 8) - (y + 2)(-5 - 12) + (z + 1)(2 + 6) = 0$
$2(x - 3) + 17(y + 2) + 8(z + 1) = 0$
$2x - 6 + 17y + 34 + 8z + 8 = 0$
$2x + 17y + 8z + 36 = 0$

(A) The intercept form 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 $a = \frac{1}{3}, b = \frac{1}{4}, c = \frac{1}{5}$.
Substituting these values,the equation of the plane is $\frac{x}{1/3} + \frac{y}{1/4} + \frac{z}{1/5} = 1$,which simplifies to $3x + 4y + 5z = 1$ or $3x + 4y + 5z - 1 = 0$.
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,$A = 3, B = 4, C = 5$ and $D = -1$.
Thus,$d = \frac{|-1|}{\sqrt{3^2 + 4^2 + 5^2}} = \frac{1}{\sqrt{9 + 16 + 25}} = \frac{1}{\sqrt{50}} = \frac{1}{5 \sqrt{2}}$.

(D) The equation of a plane passing through the point $(x_1, y_1, z_1)$ with normal vector $(a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.
Given the point $(2, 3, -1)$,the equation is $a(x-2) + b(y-3) + c(z+1) = 0 \dots (i)$.
Since the plane is perpendicular to the line with direction ratios $(3, -4, 7)$,the normal vector of the plane is parallel to this line.
Thus,we can take $a=3, b=-4, c=7$.
Substituting these into equation $(i)$:
$3(x-2) - 4(y-3) + 7(z+1) = 0$
$3x - 6 - 4y + 12 + 7z + 7 = 0$
$3x - 4y + 7z + 13 = 0$.
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,$A=3, B=-4, C=7, D=13$.
$d = \frac{|13|}{\sqrt{3^2 + (-4)^2 + 7^2}} = \frac{13}{\sqrt{9 + 16 + 49}} = \frac{13}{\sqrt{74}}$.

(B) The intercept form of the equation of a plane meeting the coordinate axes at $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$ is given by:
$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ $\ldots$ $(i)$
The centroid of the triangle $ABC$ with vertices $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$ is given by the formula:
$\left( \frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3} \right) = \left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right)$
Given that the centroid is $(1, 2, 4)$,we equate the coordinates:
$\frac{a}{3} = 1 \Rightarrow a = 3$
$\frac{b}{3} = 2 \Rightarrow b = 6$
$\frac{c}{3} = 4 \Rightarrow c = 12$
Substituting these values into Eq. $(i)$:
$\frac{x}{3} + \frac{y}{6} + \frac{z}{12} = 1$
Multiplying the entire equation by $12$ to clear the denominators:
$4x + 2y + z = 12$

(D) The given equation of the plane is $3x - 2y - z = 18$.
Dividing both sides by $18$,we get the intercept form:
$\frac{3x}{18} - \frac{2y}{18} - \frac{z}{18} = 1$
$\frac{x}{6} + \frac{y}{-9} + \frac{z}{-18} = 1$.
Comparing this with the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,the intercepts on the coordinate axes are $a = 6, b = -9, c = -18$.
Thus,the coordinates of the points where the plane meets the axes are $A(6, 0, 0)$,$B(0, -9, 0)$,and $C(0, 0, -18)$.
The centroid $(G)$ of $\triangle ABC$ is given by the formula $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
$G = \left(\frac{6+0+0}{3}, \frac{0-9+0}{3}, \frac{0+0-18}{3}\right)$
$G = (2, -3, -6)$.

(A) Let the points be $A(0,0,1)$,$B(0,1,2)$,and $C(1,0,3)$.
The vectors lying on the plane are $\vec{AB} = (0-0, 1-0, 2-1) = (0,1,1)$ and $\vec{AC} = (1-0, 0-0, 3-1) = (1,0,2)$.
The normal vector $\vec{n}$ to the plane is given by the cross product $\vec{AB} \times \vec{AC}$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 1 & 0 & 2 \end{vmatrix}$
$\vec{n} = \hat{i}(1 \times 2 - 1 \times 0) - \hat{j}(0 \times 2 - 1 \times 1) + \hat{k}(0 \times 0 - 1 \times 1)$
$\vec{n} = \hat{i}(2) - \hat{j}(-1) + \hat{k}(-1)$
$\vec{n} = 2\hat{i} + \hat{j} - \hat{k}$
The direction ratios of the normal are $(2,1,-1)$.

(A) The given equation of the plane is $by + cz + d = 0$.
Since the variable $x$ is missing from the equation,the normal vector to the plane is $\vec{n} = 0\hat{i} + b\hat{j} + c\hat{k}$.
The normal vector lies in the $YOZ$-plane.
$A$ plane is perpendicular to another plane if their normal vectors are perpendicular.
The normal to the $YOZ$-plane is the $x$-axis,which is $\hat{i}$.
Since the dot product of the normal vector $\vec{n} = b\hat{j} + c\hat{k}$ and the $x$-axis vector $\hat{i}$ is $0$,the plane $by + cz + d = 0$ is perpendicular to the $YOZ$-plane.

(C) First,we find the prime factorization of $2001$.
$2001 = 3 \times 667 = 3 \times 23 \times 29$.
Given that $a, b, c$ are factors of $2001$ such that $a < b < c$,we have $a = 3$,$b = 23$,and $c = 29$.
The direction ratios of the normal to the plane are given as $b, c, a$,which are $23, 29, 3$.
The equation of a plane with normal vector $(b, c, a)$ passing through $(x_0, y_0, z_0)$ is $b(x - x_0) + c(y - y_0) + a(z - z_0) = 0$.
Substituting the values,we get $23(x - 1) + 29(y - 1) + 3(z - 1) = 0$.
Expanding this,we get $23x - 23 + 29y - 29 + 3z - 3 = 0$.
$23x + 29y + 3z - 55 = 0$.
Therefore,the equation of the plane is $23x + 29y + 3z = 55$.

(A) The given equation of the plane is $7x + 11y + 13z = 3003$.
Dividing the entire equation by $3003$,we get the intercept form of the plane:
$\frac{7x}{3003} + \frac{11y}{3003} + \frac{13z}{3003} = 1$
$\frac{x}{429} + \frac{y}{273} + \frac{z}{231} = 1$
The plane meets the coordinate axes at points $A, B,$ and $C$. These points are the intercepts on the $x, y,$ and $z$ axes respectively:
$A = (429, 0, 0)$
$B = (0, 273, 0)$
$C = (0, 0, 231)$
The centroid of $\triangle ABC$ with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2),$ and $(x_3, y_3, z_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Centroid $= \left(\frac{429+0+0}{3}, \frac{0+273+0}{3}, \frac{0+0+231}{3}\right)$
Centroid $= (143, 91, 77)$.

(B) The normal vector $\vec{n}$ to the plane is the vector from the origin $(0,0,0)$ to the foot of the perpendicular $(1,2,2)$.
$\vec{n} = (1-0)\hat{i} + (2-0)\hat{j} + (2-0)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$.
The equation of a plane passing through a point $(x_0, y_0, z_0)$ with normal vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.
Here,$(x_0, y_0, z_0) = (1, 2, 2)$ and $(a, b, c) = (1, 2, 2)$.
Substituting these values,we get:
$1(x-1) + 2(y-2) + 2(z-2) = 0$
$x - 1 + 2y - 4 + 2z - 4 = 0$
$x + 2y + 2z - 9 = 0$.
Thus,the correct option is $B$.

(C) Since the lines $L_1$ and $L_2$ pass through the origin $(0, 0, 0)$,the plane containing them also passes through the origin. The equation of such a plane is given by $ax + by + cz = 0$.
The normal vector $\vec{n}$ to the plane is the cross product of the direction vectors of the two lines,$\vec{v}_1 = 3\hat{i} + \hat{j} - 5\hat{k}$ and $\vec{v}_2 = 2\hat{i} + 3\hat{j} - \hat{k}$.
$\vec{n} = \vec{v}_1 \times \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -5 \\ 2 & 3 & -1 \end{vmatrix}$
$\vec{n} = \hat{i}(-1 - (-15)) - \hat{j}(-3 - (-10)) + \hat{k}(9 - 2)$
$\vec{n} = 14\hat{i} - 7\hat{j} + 7\hat{k}$
Dividing by $7$,we get the normal vector as $2\hat{i} - \hat{j} + \hat{k}$.
Thus,the equation of the plane is $2(x - 0) - 1(y - 0) + 1(z - 0) = 0$,which simplifies to $2x - y + z = 0$.

(C) Plane $\pi$ passes through $(3,-3,1)$ and is perpendicular to the line joining the points $(3,4,-1)$ and $(2,-1,5)$.
The direction ratios of the normal to the plane $\pi$ are $(3-2, 4-(-1), -1-5) = (1, 5, -6)$.
Thus,the equation of plane $\pi$ is $1(x-3) + 5(y+3) - 6(z-1) = 0$,which simplifies to $x+5y-6z+18=0$.
Let the second plane be $P_2: ax+y+cz-d=0$. This plane contains the points $(3,4,-1)$ and $(-1,2,5)$.
Since $(3,4,-1)$ lies on $P_2$,we have $3a+4+c(-1)-d=0 \Rightarrow 3a-c-d=-4$ ... $(i)$.
Since $(-1,2,5)$ lies on $P_2$,we have $-a+2+5c-d=0 \Rightarrow -a+5c-d=-2$ ... (ii).
Subtracting (ii) from $(i)$,we get $4a-6c=-2 \Rightarrow 2a-3c=-1$ ... (iii).
The normal vector of $P_2$ is $\vec{n_2} = (a, 1, c)$ and the normal vector of $\pi$ is $\vec{n_1} = (1, 5, -6)$.
Since $P_2 \perp \pi$,their normals are perpendicular: $\vec{n_1} \cdot \vec{n_2} = 0 \Rightarrow a(1) + 1(5) + c(-6) = 0 \Rightarrow a-6c=-5$ ... (iv).
From (iv),$a = 6c-5$. Substituting into (iii): $2(6c-5)-3c=-1 \Rightarrow 12c-10-3c=-1 \Rightarrow 9c=9 \Rightarrow c=1$.
Then $a = 6(1)-5 = 1$.
Substituting $a=1, c=1$ into $(i)$: $3(1)-1-d=-4 \Rightarrow 2-d=-4 \Rightarrow d=6$.
Finally,$3(a+c) = 3(1+1) = 6$. Since $d=6$,$3(a+c) = d$.

(A) The mid-point $R$ of the line segment joining $P(3,2,4)$ and $Q(-1,0,-2)$ is given by:
$R = \left( \frac{3-1}{2}, \frac{2+0}{2}, \frac{4-2}{2} \right) = (1, 1, 1)$.
The direction ratios of the line segment $PQ$ are $(3 - (-1), 2 - 0, 4 - (-2)) = (4, 2, 6)$.
Since the plane is perpendicular to $PQ$,the normal vector to the plane is $\vec{n} = \langle 4, 2, 6 \rangle$.
Thus,the equation of the plane is $4x + 2y + 6z + d = 0$.
Since the plane passes through the mid-point $R(1, 1, 1)$,we have:
$4(1) + 2(1) + 6(1) + d = 0 \implies 4 + 2 + 6 + d = 0 \implies d = -12$.
Comparing $4x + 2y + 6z - 12 = 0$ with $ax + by + cz + d = 0$,we get $a=4, b=2, c=6, d=-12$.
Therefore,$ac + bd = (4)(6) + (2)(-12) = 24 - 24 = 0$.

(C) Let the points be $A(2, 0, 6)$ and $B(-6, 2, 4)$.
The plane bisects the line segment $AB$ and is perpendicular to it,meaning it passes through the midpoint of $AB$ and the vector $\vec{AB}$ is the normal to the plane.
The midpoint $M$ of $AB$ is $\left(\frac{2-6}{2}, \frac{0+2}{2}, \frac{6+4}{2}\right) = (-2, 1, 5)$.
The normal vector $\vec{n}$ to the plane is $\vec{AB} = (-6-2, 2-0, 4-6) = (-8, 2, -2)$.
We can simplify the normal vector by dividing by $-2$,giving $\vec{n}' = (4, -1, 1)$.
The equation of the plane passing through $(x_0, y_0, z_0)$ with normal $(a, b, c)$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.
Substituting the values: $4(x - (-2)) - 1(y - 1) + 1(z - 5) = 0$.
$4(x+2) - y + 1 + z - 5 = 0$.
$4x + 8 - y + z - 4 = 0$.
$4x - y + z + 4 = 0$.

(B) The plane is perpendicular to the line segment $AB$ and passes through its midpoint.
First,find the midpoint $M$ of the line segment $AB$:
$M = \left( \frac{2 + (-4)}{2}, \frac{3 + 1}{2}, \frac{4 + (-2)}{2} \right) = (-1, 2, 1)$.
Next,find the normal vector $\vec{n}$ to the plane,which is the vector $\vec{AB}$:
$\vec{n} = \vec{AB} = (-4 - 2, 1 - 3, -2 - 4) = (-6, -2, -6)$.
We can simplify the normal vector by dividing by $-2$:
$\vec{n}' = (3, 1, 3)$.
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:
$3(x - (-1)) + 1(y - 2) + 3(z - 1) = 0$
$3(x + 1) + (y - 2) + 3(z - 1) = 0$
$3x + 3 + y - 2 + 3z - 3 = 0$
$3x + y + 3z - 2 = 0$.
Thus,the correct option is $B$.

(B) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ ... $(i)$
Since the plane meets the coordinate axes at $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$,the centroid of $\triangle ABC$ 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:
$\frac{a}{3} = 1 \Rightarrow a = 3$
$\frac{b}{3} = 2 \Rightarrow b = 6$
$\frac{c}{3} = 4 \Rightarrow c = 12$
Substituting these values into equation $(i)$,we get:
$\frac{x}{3} + \frac{y}{6} + \frac{z}{12} = 1$
Multiplying by $12$,we get $4x + 2y + z = 12$.

(A) Let $P$ be the point nearest to $A = \left(1, \frac{3}{2}, 2\right)$ on the plane $2x - 2y + 4z + 5 = 0$.
The normal vector to the plane is $\vec{n} = (2, -2, 4)$.
The line passing through $A$ and perpendicular to the plane is given by $P = A + c\vec{n} = \left(1 + 2c, \frac{3}{2} - 2c, 2 + 4c\right)$.
Since $P$ lies on the plane,substitute the coordinates into the plane equation:
$2(1 + 2c) - 2\left(\frac{3}{2} - 2c\right) + 4(2 + 4c) + 5 = 0$.
$2 + 4c - 3 + 4c + 8 + 16c + 5 = 0$.
$24c + 12 = 0 \Rightarrow c = -\frac{1}{2}$.
Substituting $c = -\frac{1}{2}$ back into the expression for $P$:
$P = \left(1 + 2(-\frac{1}{2}), \frac{3}{2} - 2(-\frac{1}{2}), 2 + 4(-\frac{1}{2})\right) = \left(0, \frac{5}{2}, 0\right)$.

(D) The normals to the given planes $4x + 3y - 5 = 0$ and $x + 2y + 2z - 4 = 0$ are $\vec{n_1} = 4\hat{i} + 3\hat{j} + 0\hat{k}$ and $\vec{n_2} = \hat{i} + 2\hat{j} + 2\hat{k}$.
To find the bisector of the angle between these normals,we first normalize the vectors:
$|\vec{n_1}| = \sqrt{4^2 + 3^2 + 0^2} = 5$
$|\vec{n_2}| = \sqrt{1^2 + 2^2 + 2^2} = 3$
The unit vectors are $\hat{n_1} = \frac{4}{5}\hat{i} + \frac{3}{5}\hat{j}$ and $\hat{n_2} = \frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$.
The angle bisector is along the vector $\hat{n_1} \pm \hat{n_2}$.
Taking the sum: $(\frac{4}{5} + \frac{1}{3})\hat{i} + (\frac{3}{5} + \frac{2}{3})\hat{j} + \frac{2}{3}\hat{k} = \frac{17}{15}\hat{i} + \frac{19}{15}\hat{j} + \frac{10}{15}\hat{k}$,which is proportional to $17\hat{i} + 19\hat{j} + 10\hat{k}$.
Taking the difference: $(\frac{4}{5} - \frac{1}{3})\hat{i} + (\frac{3}{5} - \frac{2}{3})\hat{j} - \frac{2}{3}\hat{k} = \frac{7}{15}\hat{i} - \frac{1}{15}\hat{j} - \frac{10}{15}\hat{k}$,which is proportional to $7\hat{i} - \hat{j} - 10\hat{k}$.
Comparing with the options,$7\hat{i} - \hat{j} - 10\hat{k}$ is the correct vector.

(D) The equations of the two planes are given by $\bar{r} \cdot \vec{n}_1 = d_1$ and $\bar{r} \cdot \vec{n}_2 = d_2$,where $\vec{n}_1 = 11 \hat{i} - 2 \hat{j} + \alpha \hat{k}$ and $\vec{n}_2 = 2 \hat{i} + 4 \hat{j} - 2 \hat{k}$.
Since the angle between the planes is $\frac{\pi}{2}$,the planes are perpendicular to each other.
For two perpendicular planes,the dot product of their normal vectors must be zero,i.e.,$\vec{n}_1 \cdot \vec{n}_2 = 0$.
Substituting the vectors,we get $(11 \hat{i} - 2 \hat{j} + \alpha \hat{k}) \cdot (2 \hat{i} + 4 \hat{j} - 2 \hat{k}) = 0$.
Calculating the dot product: $(11 \times 2) + (-2 \times 4) + (\alpha \times -2) = 0$.
$22 - 8 - 2\alpha = 0$.
$14 - 2\alpha = 0$.
$2\alpha = 14$.
$\alpha = 7$.

(C) Let the points be $A(1, -2, -3)$,$B(3, -1, 4)$,and $C(-3, 2, -5)$.
First,we find two vectors in the plane: $\vec{AB} = (3-1)\hat{i} + (-1+2)\hat{j} + (4+3)\hat{k} = 2\hat{i} + \hat{j} + 7\hat{k}$ and $\vec{AC} = (-3-1)\hat{i} + (2+2)\hat{j} + (-5+3)\hat{k} = -4\hat{i} + 4\hat{j} - 2\hat{k}$.
The normal vector $\vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 7 \\ -4 & 4 & -2 \end{vmatrix} = \hat{i}(-2-28) - \hat{j}(-4+28) + \hat{k}(8+4) = -30\hat{i} - 24\hat{j} + 12\hat{k}$.
Dividing by $-6$,we get the normal vector $\vec{n}' = 5\hat{i} + 4\hat{j} - 2\hat{k}$.
The equation of the plane is $5(x-1) + 4(y+2) - 2(z+3) = 0$,which simplifies to $5x + 4y - 2z - 5 + 8 - 6 = 0$,or $5x + 4y - 2z - 3 = 0$.
Checking option $C$: $5(-1) + 4(9) - 2(14) - 3 = -5 + 36 - 28 - 3 = 0$. Thus,the point $-\hat{i} + 9\hat{j} + 14\hat{k}$ lies on the plane.

(D) Given the vector equation of the plane is $r=(2+3 \lambda-\mu) \hat{i}+(1-2 \lambda+3 \mu) \hat{j}+(-2+2 \lambda+\mu) \hat{k}$ $\ldots$ $(i)$
Let $r=x \hat{i}+y \hat{j}+z \hat{k}$ $\ldots$ (ii)
Comparing components from $(i)$ and (ii):
$x = 2+3 \lambda-\mu$ $\ldots$ (iii)
$y = 1-2 \lambda+3 \mu$ $\ldots$ (iv)
$z = -2+2 \lambda+\mu$ $\ldots$ $(v)$
Adding (iii) and $(v)$: $x+z = 2-2+3 \lambda+2 \lambda-\mu+\mu = 5 \lambda \Rightarrow \lambda = \frac{x+z}{5}$ $\ldots$ (vi)
From $(v)$,$\mu = z+2-2 \lambda = z+2-2(\frac{x+z}{5}) = \frac{5z+10-2x-2z}{5} = \frac{-2x+3z+10}{5}$
Substitute $\lambda$ and $\mu$ into (iv): $y = 1-2(\frac{x+z}{5})+3(\frac{-2x+3z+10}{5})$
$5y = 5-2x-2z-6x+9z+30$
$5y = -8x+7z+35$
$8x+5y-7z-35=0$

(B) The given vector equation is $\overrightarrow{r}=(1-p-q) \overrightarrow{a}+p \overrightarrow{b}+q \overrightarrow{c}$.
We can rewrite this as $\overrightarrow{r} = \overrightarrow{a} - p\overrightarrow{a} - q\overrightarrow{a} + p\overrightarrow{b} + q\overrightarrow{c}$.
Rearranging the terms,we get $\overrightarrow{r} - \overrightarrow{a} = p(\overrightarrow{b} - \overrightarrow{a}) + q(\overrightarrow{c} - \overrightarrow{a})$.
This is the parametric form of the equation of a plane passing through the point with position vector $\overrightarrow{a}$ and parallel to the vectors $(\overrightarrow{b} - \overrightarrow{a})$ and $(\overrightarrow{c} - \overrightarrow{a})$.
Since $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are non-coplanar,the vectors $(\overrightarrow{b} - \overrightarrow{a})$ and $(\overrightarrow{c} - \overrightarrow{a})$ are linearly independent,thus defining a unique plane.

(C) The equation of the plane is $3x + 4y - 5z = 60$.
Dividing by $60$,we get the intercept form: $\frac{x}{20} + \frac{y}{15} - \frac{z}{12} = 1$.
This plane intersects the coordinate axes at the points $A(20, 0, 0)$,$B(0, 15, 0)$,and $C(0, 0, -12)$.
The tetrahedron is formed by the origin $O(0, 0, 0)$ and the points $A, B, C$.
The volume of a tetrahedron with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3)$ and $(x_4, y_4, z_4)$ is given by $V = \frac{1}{6} |\vec{OA} \cdot (\vec{OB} \times \vec{OC})|$.
Here,$\vec{OA} = 20\hat{i}$,$\vec{OB} = 15\hat{j}$,and $\vec{OC} = -12\hat{k}$.
$V = \frac{1}{6} |20\hat{i} \cdot (15\hat{j} \times -12\hat{k})| = \frac{1}{6} |20 \times 15 \times (-12)| = \frac{1}{6} |-3600| = 600$ cubic units.

(D) The equation of the given plane is $2x + y + z = K$.
Dividing by $K$,we get the intercept form: $\frac{x}{K/2} + \frac{y}{K} + \frac{z}{K} = 1$.
The vertices of the tetrahedron formed by the plane and the coordinate planes are $O(0, 0, 0)$,$A(K/2, 0, 0)$,$B(0, K, 0)$,and $C(0, 0, K)$.
The volume of the tetrahedron $OABC$ is given by the formula $V_{tet} = \frac{1}{6} |x_A y_B z_C| = \frac{1}{6} \times \frac{K}{2} \times K \times K = \frac{K^3}{12}$.
Given that the volume is $\frac{2V^3}{3}$,we equate the two expressions:
$\frac{K^3}{12} = \frac{2V^3}{3}$.
Multiplying both sides by $12$,we get $K^3 = 8V^3$.
Taking the cube root on both sides,we get $K = 2V$,which implies $\frac{K}{V} = \frac{2}{1}$.
Thus,$K:V = 2:1$.
Therefore,option $D$ is correct.

(A) The normal vectors to the planes $ax - y + 3z = 2a$ and $3x + ay + z = 3a$ are $\vec{n_1} = (a, -1, 3)$ and $\vec{n_2} = (3, a, 1)$ respectively.
Given the angle $\theta = \frac{\pi}{3}$ between the planes,we have $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
$\cos(\frac{\pi}{3}) = \frac{|3a - a + 3|}{\sqrt{a^2 + (-1)^2 + 3^2} \sqrt{3^2 + a^2 + 1^2}} = \frac{1}{2}$.
$\frac{|2a + 3|}{\sqrt{a^2 + 10} \sqrt{a^2 + 10}} = \frac{1}{2} \implies \frac{|2a + 3|}{a^2 + 10} = \frac{1}{2}$.
$2|2a + 3| = a^2 + 10$.
Case $1$: $4a + 6 = a^2 + 10 \implies a^2 - 4a + 4 = 0 \implies (a - 2)^2 = 0 \implies a = 2$.
Case $2$: $-4a - 6 = a^2 + 10 \implies a^2 + 4a + 16 = 0$,which has no real roots.
Thus,$a = 2$.
The plane equation becomes $(2+2)x + (2-4)y + 2(2)z = 2$,which is $4x - 2y + 4z = 2$,or $2x - y + 2z = 1$.
The direction ratios of the normal to this plane are $(2, -1, 2)$.

(A) Let the two vectors be $\vec{b_1} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b_2} = 4\hat{i} + 5\hat{j} - 3\hat{k}$.
The normal vector $\vec{n}$ to the plane containing these lines is given by the cross product $\vec{b_1} \times \vec{b_2}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 4 & 5 & -3 \end{vmatrix} = \hat{i}(-6 - 5) - \hat{j}(-3 - 4) + \hat{k}(5 - 8) = -11\hat{i} + 7\hat{j} - 3\hat{k}$.
The magnitude of the normal vector is $|\vec{n}| = \sqrt{(-11)^2 + 7^2 + (-3)^2} = \sqrt{121 + 49 + 9} = \sqrt{179}$.
The direction cosines are given by $\frac{a}{|\vec{n}|}, \frac{b}{|\vec{n}|}, \frac{c}{|\vec{n}|}$,where $a, b, c$ are the components of the normal vector.
Thus,the direction cosines are $\frac{-11}{\sqrt{179}}, \frac{7}{\sqrt{179}}, \frac{-3}{\sqrt{179}}$.

(A) Let the direction ratios of the lines $AB$ and $AC$ be $\vec{u} = \langle 1, -1, -1 \rangle$ and $\vec{v} = \langle 2, -1, 1 \rangle$ respectively.
Since the lines $AB$ and $AC$ lie in the plane $ABC$,their cross product $\vec{n} = \vec{u} \times \vec{v}$ will give the direction ratios of the normal to the plane.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -1 \\ 2 & -1 & 1 \end{vmatrix}$
$= \hat{i}(-1 - 1) - \hat{j}(1 - (-2)) + \hat{k}(-1 - (-2))$
$= \hat{i}(-2) - \hat{j}(3) + \hat{k}(1)$
$= -2\hat{i} - 3\hat{j} + \hat{k}$
Thus,the direction ratios are $\langle -2, -3, 1 \rangle$.
Multiplying by $-1$,we get the equivalent direction ratios $\langle 2, 3, -1 \rangle$.
Therefore,the correct option is $A$.

(D) The normal vector $\vec{n}$ to the plane is given by the cross product of vectors $\vec{AB}$ and $\vec{AC}$.
$\vec{AB} = (6-2, -3-1, 2-(-1)) = (4, -4, 3)$.
$\vec{AC} = (-3-2, 12-1, 4-(-1)) = (-5, 11, 5)$.
$\vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -4 & 3 \\ -5 & 11 & 5 \end{vmatrix} = \hat{i}(-20-33) - \hat{j}(20+15) + \hat{k}(44-20) = -53\hat{i} - 35\hat{j} + 24\hat{k}$.
The equation of the plane is $-53(x-2) - 35(y-1) + 24(z+1) = 0$.
$-53x + 106 - 35y + 35 + 24z + 24 = 0$.
$-53x - 35y + 24z + 165 = 0$.
Multiply by $-1$ to match the form $53x + by + cz + d = 0$:
$53x + 35y - 24z - 165 = 0$.
Here,$b = 35$,$c = -24$,and $d = -165$.
Then,$\frac{d}{b+c} = \frac{-165}{35 - 24} = \frac{-165}{11} = -15$.

(C) The normal vector $\vec{n}$ to the plane is the vector $\vec{PF}$.
$\vec{PF} = (1-2, -2-0, 0-(-3)) = (-1, -2, 3)$.
Since the equation of the plane is $ax+by-3z+d=0$,the normal vector is $(a, b, -3)$.
Comparing $(a, b, -3)$ with $k(-1, -2, 3)$,we get $k(-1) = a$,$k(-2) = b$,and $k(3) = -3$.
Thus,$k = -1$.
Therefore,$a = -1(-1) = 1$ and $b = -1(-2) = 2$.
The equation of the plane is $x+2y-3z+d=0$.
Since the point $F(1,-2,0)$ lies on the plane,we have $1 + 2(-2) - 3(0) + d = 0$.
$1 - 4 + d = 0 \implies -3 + d = 0 \implies d = 3$.
We need to find $a+b+d = 1 + 2 + 3 = 6$.

(C) The given planes are $P_1: -4x - 2y + 2z + \alpha = 0$ and $P_2: 2x + y - z + 1 = 0$.
First,rewrite $P_1$ by dividing by $-2$: $2x + y - z - \frac{\alpha}{2} = 0$.
Let $k = -\frac{\alpha}{2}$. The planes are $2x + y - z + k = 0$ and $2x + y - z + 1 = 0$.
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,$d = 2$,$A = 2$,$B = 1$,$C = -1$,$D_1 = k$,and $D_2 = 1$.
$2 = \frac{|k - 1|}{\sqrt{2^2 + 1^2 + (-1)^2}} = \frac{|k - 1|}{\sqrt{6}}$.
$|k - 1| = 2\sqrt{6}$.
$k - 1 = 2\sqrt{6}$ or $k - 1 = -2\sqrt{6}$.
$k = 1 + 2\sqrt{6}$ or $k = 1 - 2\sqrt{6}$.
Since $k = -\frac{\alpha}{2}$,we have $\alpha = -2k$.
$\alpha_1 = -2(1 + 2\sqrt{6}) = -2 - 4\sqrt{6}$ and $\alpha_2 = -2(1 - 2\sqrt{6}) = -2 + 4\sqrt{6}$.
The product of the values of $\alpha$ is $\alpha_1 \alpha_2 = (-2 - 4\sqrt{6})(-2 + 4\sqrt{6}) = (-2)^2 - (4\sqrt{6})^2 = 4 - 16(6) = 4 - 96 = -92$.

(C) Let the points be $A(5,1,2)$,$B(3,-4,6)$,and $C(7,0,-1)$.
The equation of the plane passing through these points is given by $\begin{vmatrix} x-5 & y-1 & z-2 \\ 3-5 & -4-1 & 6-2 \\ 7-5 & 0-1 & -1-2 \end{vmatrix} = 0$.
$\begin{vmatrix} x-5 & y-1 & z-2 \\ -2 & -5 & 4 \\ 2 & -1 & -3 \end{vmatrix} = 0$.
$(x-5)(15+4) - (y-1)(6-8) + (z-2)(2+10) = 0$.
$19(x-5) + 2(y-1) + 12(z-2) = 0$.
$19x - 95 + 2y - 2 + 12z - 24 = 0$.
$19x + 2y + 12z - 121 = 0$.
The normal vector is $\vec{n} = 19\hat{i} + 2\hat{j} + 12\hat{k}$.
The magnitude is $|\vec{n}| = \sqrt{19^2 + 2^2 + 12^2} = \sqrt{361 + 4 + 144} = \sqrt{509}$.
The direction cosines are $l = \frac{19}{\sqrt{509}}$,$m = \frac{2}{\sqrt{509}}$,$n = \frac{12}{\sqrt{509}}$.
The perpendicular distance $p$ from the origin $(0,0,0)$ to the plane $19x + 2y + 12z - 121 = 0$ is $p = \frac{|-121|}{\sqrt{509}} = \frac{121}{\sqrt{509}}$.
Now,$|3l + 2m + 5n| = |3(\frac{19}{\sqrt{509}}) + 2(\frac{2}{\sqrt{509}}) + 5(\frac{12}{\sqrt{509}})| = |\frac{57 + 4 + 60}{\sqrt{509}}| = \frac{121}{\sqrt{509}} = p$.

(C) Let the equation of the plane be $\pi: ax + by + cz + 1 = 0$.
Since it passes through $(1, 2, -3)$,we have $a(1) + b(2) + c(-3) + 1 = 0$,which implies $a + 2b - 3c = -1$ (Equation $i$).
Since the plane is perpendicular to $x + y - z + 4 = 0$ and $2x - y + z + 1 = 0$,its normal vector $\vec{n} = (a, b, c)$ is perpendicular to the normal vectors of these planes,$\vec{n_1} = (1, 1, -1)$ and $\vec{n_2} = (2, -1, 1)$.
Thus,$a + b - c = 0$ (Equation $ii$) and $2a - b + c = 0$ (Equation $iii$).
Adding equations $(ii)$ and $(iii)$,we get $3a = 0$,so $a = 0$.
Substituting $a = 0$ into $(ii)$,we get $b - c = 0$,so $b = c$.
Substituting $a = 0$ and $b = c$ into $(i)$,we get $0 + 2c - 3c = -1$,which implies $-c = -1$,so $c = 1$.
Thus,$b = 1$.
The values are $a = 0, b = 1, c = 1$.
Therefore,$a^2 + b^2 + c^2 = 0^2 + 1^2 + 1^2 = 2$.

(C) 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 given points $(1, 0, -2), (3, -1, 2)$ and $(0, -3, 4)$:
$\begin{vmatrix} x-1 & y-0 & z+2 \\ 3-1 & -1-0 & 2+2 \\ 0-1 & -3-0 & 4+2 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} x-1 & y & z+2 \\ 2 & -1 & 4 \\ -1 & -3 & 6 \end{vmatrix} = 0$
Expanding the determinant:
$(x-1)(-6 + 12) - y(12 + 4) + (z+2)(-6 - 1) = 0$
$6(x-1) - 16y - 7(z+2) = 0$
$6x - 6 - 16y - 7z - 14 = 0 \Rightarrow 6x - 16y - 7z = 20$
Dividing by $20$ to get the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$:
$\frac{6x}{20} - \frac{16y}{20} - \frac{7z}{20} = 1 \Rightarrow \frac{x}{10/3} + \frac{y}{-20/16} + \frac{z}{-20/7} = 1$
Thus,$a = \frac{10}{3}, b = -\frac{5}{4}, c = -\frac{20}{7}$
Calculating $3a + 4b + 7c$:
$3(\frac{10}{3}) + 4(-\frac{5}{4}) + 7(-\frac{20}{7}) = 10 - 5 - 20 = -15$.

(C) The vector equation of a plane passing through point $\vec{a}$ with normal vector $\vec{n}$ is $(\vec{r}-\vec{a}) \cdot \vec{n} = 0$,which simplifies to $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
For plane $\pi_1$,$\vec{a}_1 = 3 \hat{i}-7 \hat{j}+5 \hat{k}$ and $\vec{n}_1 = \hat{i}+2 \hat{j}-2 \hat{k}$.
$\vec{r} \cdot (\hat{i}+2 \hat{j}-2 \hat{k}) = (3 \hat{i}-7 \hat{j}+5 \hat{k}) \cdot (\hat{i}+2 \hat{j}-2 \hat{k}) = 3 - 14 - 10 = -21$.
The normal form is $\vec{r} \cdot \hat{n} = p$. Here,$|\vec{n}_1| = \sqrt{1^2+2^2+(-2)^2} = 3$.
Dividing by $3$,$\vec{r} \cdot \frac{\hat{i}+2 \hat{j}-2 \hat{k}}{3} = \frac{-21}{3} = -7$. Thus,$p_1 = |-7| = 7$.
For plane $\pi_2$,$\vec{a}_2 = 2 \hat{i}+7 \hat{j}-8 \hat{k}$ and $\vec{n}_2 = 3 \hat{i}+2 \hat{j}+6 \hat{k}$.
$\vec{r} \cdot (3 \hat{i}+2 \hat{j}+6 \hat{k}) = (2 \hat{i}+7 \hat{j}-8 \hat{k}) \cdot (3 \hat{i}+2 \hat{j}+6 \hat{k}) = 6 + 14 - 48 = -28$.
Here,$|\vec{n}_2| = \sqrt{3^2+2^2+6^2} = \sqrt{9+4+36} = 7$.
Dividing by $7$,$\vec{r} \cdot \frac{3 \hat{i}+2 \hat{j}+6 \hat{k}}{7} = \frac{-28}{7} = -4$. Thus,$p_2 = |-4| = 4$.
Therefore,$p_1 - p_2 = 7 - 4 = 3$.

(B) The equation of the plane $\pi$ containing two vectors $\vec{\alpha}$ and $\vec{\beta}$ and passing through point $\vec{a}$ is given by $(\vec{r}-\vec{a}) \cdot (\vec{\alpha} \times \vec{\beta}) = 0$.
First,calculate the normal vector $\vec{n}' = \vec{\alpha} \times \vec{\beta}$:
$\vec{n}' = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 3 \\ 2 & 1 & -1 \end{vmatrix} = \hat{i}(0-3) - \hat{j}(-1-6) + \hat{k}(1-0) = -3\hat{i} + 7\hat{j} + \hat{k}$.
The equation of the plane is $-3(x+3) + 7(y-7) + 1(z-1) = 0$.
Simplifying this,we get $-3x - 9 + 7y - 49 + z - 1 = 0$,which is $-3x + 7y + z - 59 = 0$.
The perpendicular distance $p$ from the origin $(0,0,0)$ to the plane $Ax+By+Cz+D=0$ is $p = \frac{|D|}{\sqrt{A^2+B^2+C^2}}$.
$p = \frac{|-59|}{\sqrt{(-3)^2 + 7^2 + 1^2}} = \frac{59}{\sqrt{9+49+1}} = \frac{59}{\sqrt{59}} = \sqrt{59}$.
Therefore,$\sqrt{p^2+5} = \sqrt{(\sqrt{59})^2 + 5} = \sqrt{59+5} = \sqrt{64} = 8$.

(B) The direction ratios ($D$.$R$.) of the line segment $AB$ are given by $(-1-4, 2-5, 1-(-10)) = (-5, -3, 11)$.
Since the plane is perpendicular to $AB$,the normal vector to the plane is $\vec{n} = (-5, -3, 11)$.
The midpoint $P$ of $AB$ is $\left(\frac{4-1}{2}, \frac{5+2}{2}, \frac{-10+1}{2}\right) = \left(\frac{3}{2}, \frac{7}{2}, -\frac{9}{2}\right)$.
The equation of the plane passing through $(x_0, y_0, z_0)$ with normal $(a, b, c)$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.
Substituting the values: $-5(x - \frac{3}{2}) - 3(y - \frac{7}{2}) + 11(z + \frac{9}{2}) = 0$.
Multiplying by $2$: $-5(2x - 3) - 3(2y - 7) + 11(2z + 9) = 0$.
$-10x + 15 - 6y + 21 + 22z + 99 = 0$.
$-10x - 6y + 22z + 135 = 0$.
Multiplying by $-1$: $10x + 6y - 22z - 135 = 0$.

(D) The equation of the plane passing through $A(3,-2,1)$ and perpendicular to the normal vector $\vec{n} = 4 \hat{i}+7 \hat{j}-4 \hat{k}$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.
Substituting the values,we get $4(x-3)+7(y+2)-4(z-1)=0$,which simplifies to $4x+7y-4z+6=0$.
The length of the perpendicular from a point $(x_1, y_1, z_1)$ to a plane $ax+by+cz+d=0$ is given by the formula $d = \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$.
Here,the point is $P(1,2,-1)$ and the plane is $4x+7y-4z+6=0$.
Substituting these values into the formula:
$PM = \frac{|4(1)+7(2)-4(-1)+6|}{\sqrt{4^2+7^2+(-4)^2}}$
$PM = \frac{|4+14+4+6|}{\sqrt{16+49+16}}$
$PM = \frac{|28|}{\sqrt{81}}$
$PM = \frac{28}{9}$.

(D) Since the plane $\pi$ is parallel to the plane $x-y-z=0$,its equation must be of the form $x-y-z+k=0$.
Given that the point $(1, -2, 1)$ lies on the plane $\pi$,we substitute these coordinates into the equation:
$1 - (-2) - 1 + k = 0$
$1 + 2 - 1 + k = 0$
$2 + k = 0 \implies k = -2$.
Thus,the equation of the plane $\pi$ is $x-y-z-2=0$.
Comparing this with the given equation $ax+by+cz-2=0$,we get $a=1, b=-1, c=-1$.
Now,calculating $b-2c$:
$b-2c = -1 - 2(-1) = -1 + 2 = 1$.
Since $a=1$,we have $b-2c = a$.