Plane Questions in English

(D) The equation of the first plane is $ax + by + 0z + d = 0$. The normal vector to this plane is $\vec{n_1} = (a, b, 0)$.
The equation of the second plane is $0x + 0y + 1z = 0$. The normal vector to this plane is $\vec{n_2} = (0, 0, 1)$.
The angle $\theta$ between two planes is given by $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
Calculating the dot product: $\vec{n_1} \cdot \vec{n_2} = (a)(0) + (b)(0) + (0)(1) = 0$.
Since the dot product is $0$,$\cos \theta = 0$,which implies $\theta = \frac{\pi}{2}$.

(B) Let the moving point be $P(x, y, z)$.
According to the given condition,the distance from $P$ to $A(3, 4, -2)$ is equal to the distance from $P$ to $B(2, 3, -3)$.
So,$PA^2 = PB^2$.
$(x - 3)^2 + (y - 4)^2 + (z + 2)^2 = (x - 2)^2 + (y - 3)^2 + (z + 3)^2$
Expanding both sides:
$(x^2 - 6x + 9) + (y^2 - 8y + 16) + (z^2 + 4z + 4) = (x^2 - 4x + 4) + (y^2 - 6y + 9) + (z^2 + 6z + 9)$
Canceling $x^2, y^2, z^2$ from both sides:
$-6x - 8y + 4z + 29 = -4x - 6y + 6z + 22$
Rearranging the terms:
$2x + 2y + 2z = 7$
This is the equation of a plane. The normal vector to this plane is $\vec{n} = (2, 2, 2)$.
The direction cosines of the normal are proportional to $(1, 1, 1)$,which means the normal makes equal angles with the coordinate axes.

(C) The intercept form of the equation of a plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b$,and $c$ are the intercepts on the $x, y$,and $z$ axes respectively.
Given intercepts are $a = -4, b = 2$,and $c = 3$.
Substituting these values into the formula:
$\frac{x}{-4} + \frac{y}{2} + \frac{z}{3} = 1$
To simplify,find the least common multiple of the denominators $4, 2$,and $3$,which is $12$.
Multiplying the entire equation by $12$:
$12 \times (\frac{x}{-4}) + 12 \times (\frac{y}{2}) + 12 \times (\frac{z}{3}) = 12 \times 1$
$-3x + 6y + 4z = 12$
Thus,the equation of the plane is $-3x + 6y + 4z = 12$.

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

(B) Let the equation of the plane be $a(x - 1) + by + cz = 0$,which simplifies to $ax + by + cz = a$. Since it passes through $(0, 1, 0)$,we have $a(0) + b(1) + c(0) = a$,so $b = a$. The equation becomes $ax + ay + cz = a$,or $x + y + (c/a)z = 1$. Let $k = c/a$. The normal vector is $\vec{n_1} = (1, 1, k)$. The normal to the plane $x + y = 3$ is $\vec{n_2} = (1, 1, 0)$. The angle between the planes is $\pi/4$,so $\cos(\pi/4) = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
$1/\sqrt{2} = \frac{|1+1+0|}{\sqrt{1^2+1^2+k^2} \sqrt{1^2+1^2+0^2}} = \frac{2}{\sqrt{2+k^2} \sqrt{2}}$.
Squaring both sides: $1/2 = \frac{4}{2(2+k^2)} = \frac{2}{2+k^2}$.
$2+k^2 = 4 \implies k^2 = 2 \implies k = \pm \sqrt{2}$.
Taking $k = \sqrt{2}$,the normal is $(1, 1, \sqrt{2})$. Thus,the direction ratios are proportional to $1, 1, \sqrt{2}$.

(C) The normal vectors to the planes are:
$\vec{n}_P = (1, 1, -2)$
$\vec{n}_Q = (1, 1, 2)$
$\vec{n}_R = (3, 3, -6)$
Two planes are parallel if their normal vectors are proportional.
For $P$ and $R$,we check the ratio of components:
$\frac{1}{3} = \frac{1}{3} = \frac{-2}{-6} = \frac{1}{3}$.
Since the normal vectors $\vec{n}_P$ and $\vec{n}_R$ are proportional,the planes $P$ and $R$ are parallel.
Therefore,the correct option is $C$.

(A) The given line is $\vec{r} = (2, 3, 4) + k(3, 4, 5)$.
The direction vector of this line is $\vec{v} = (3, 4, 5)$.
Since the plane is perpendicular to the line,the direction vector of the line acts as the normal vector $\vec{n}$ to the plane.
Thus,$\vec{n} = (3, 4, 5)$.
The plane passes through the point $(x_0, y_0, z_0) = (1, 1, 0)$.
The equation of a plane passing through $(x_0, y_0, z_0)$ with normal vector $(a, b, c)$ is given by $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.
Substituting the values: $3(x - 1) + 4(y - 1) + 5(z - 0) = 0$.
Expanding this: $3x - 3 + 4y - 4 + 5z = 0$.
Simplifying: $3x + 4y + 5z = 7$.

(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 $A, B,$ and $C$,the coordinates are $A(a, 0, 0)$,$B(0, b, 0)$,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 $(1, 2, 3)$,we have $\frac{a}{3} = 1$,$\frac{b}{3} = 2$,and $\frac{c}{3} = 3$.
This gives $a = 3$,$b = 6$,and $c = 9$.
Substituting these values into equation $(i)$,we get $\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1$.

(D) The given plane is $\frac{x}{2} + \frac{y}{3} + \frac{z}{3} = 1$.
The intercepts on the axes are $a=2, b=3, c=3$.
Thus,the coordinates of the vertices are $A(2, 0, 0)$,$B(0, 3, 0)$,and $C(0, 0, 3)$.
The vectors forming the sides are $\vec{AB} = B - A = (-2, 3, 0)$ and $\vec{AC} = C - A = (-2, 0, 3)$.
The area of $\Delta ABC$ is given by $\frac{1}{2} |\vec{AB} \times \vec{AC}|$.
Calculating the cross product: $\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & 0 \\ -2 & 0 & 3 \end{vmatrix} = \hat{i}(9 - 0) - \hat{j}(-6 - 0) + \hat{k}(0 - (-6)) = 9\hat{i} + 6\hat{j} + 6\hat{k}$.
The magnitude is $|\vec{AB} \times \vec{AC}| = \sqrt{9^2 + 6^2 + 6^2} = \sqrt{81 + 36 + 36} = \sqrt{153} = 3\sqrt{17}$.
Therefore,the area is $\frac{1}{2} \times 3\sqrt{17} = \frac{3\sqrt{17}}{2}$ sq. units.
Since this value is not among the options,the correct choice is $D$.

(D) The equation of a plane parallel to the plane $\vec{r} \cdot \vec{n} = d$ is given by $\vec{r} \cdot \vec{n} = d_1$.
Given the plane $\vec{r} \cdot (4\hat{i} - 12\hat{j} - 3\hat{k}) = 7$,the normal vector is $\vec{n} = 4\hat{i} - 12\hat{j} - 3\hat{k}$.
The equation of the required plane is $\vec{r} \cdot (4\hat{i} - 12\hat{j} - 3\hat{k}) = d_1$.
Since the plane passes through the point $\vec{a} = 2\hat{i} + \hat{j} - 4\hat{k}$,we substitute $\vec{r} = \vec{a}$ into the equation:
$d_1 = (2\hat{i} + \hat{j} - 4\hat{k}) \cdot (4\hat{i} - 12\hat{j} - 3\hat{k})$
$d_1 = (2)(4) + (1)(-12) + (-4)(-3)$
$d_1 = 8 - 12 + 12 = 8$.
Thus,the equation of the plane is $\vec{r} \cdot (4\hat{i} - 12\hat{j} - 3\hat{k}) = 8$.

(B) The equation of any plane passing through the intersection of the planes $P_1: x + 2y + 3z - 4 = 0$ and $P_2: 4x + 3y + 2z + 1 = 0$ is given by $P_1 + \lambda P_2 = 0$.
$(x + 2y + 3z - 4) + \lambda (4x + 3y + 2z + 1) = 0$.
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:
$(x + 2y + 3z - 4) + 4(4x + 3y + 2z + 1) = 0$.
$x + 2y + 3z - 4 + 16x + 12y + 8z + 4 = 0$.
$17x + 14y + 11z = 0$.

(D) Let the equation of the plane be $f(x, y, z) = 3x + 4y - 12z + 13 = 0$.
Two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ lie on opposite sides of a plane if $f(x_1, y_1, z_1)$ and $f(x_2, y_2, z_2)$ have opposite signs,i.e.,$f(x_1, y_1, z_1) \times f(x_2, y_2, z_2) < 0$.
For point $P_1 = (1, a, 1)$,$f(1, a, 1) = 3(1) + 4(a) - 12(1) + 13 = 3 + 4a - 12 + 13 = 4a + 4$.
For point $P_2 = (-3, 0, a)$,$f(-3, 0, a) = 3(-3) + 4(0) - 12(a) + 13 = -9 - 12a + 13 = 4 - 12a$.
We require $(4a + 4)(4 - 12a) < 0$.
Dividing by $16$,we get $(a + 1)(1 - 3a) < 0$.
Multiplying by $-1$ reverses the inequality: $(a + 1)(3a - 1) > 0$.
The roots are $a = -1$ and $a = 1/3$.
The inequality $(a + 1)(3a - 1) > 0$ holds when $a < -1$ or $a > 1/3$.

(A) The given equation is $3y + 4z = 0$.
This is a linear equation in three variables $x, y, z$ of the form $Ax + By + Cz + D = 0$,where $A=0, B=3, C=4$,and $D=0$.
Since $D=0$,the plane passes through the origin $(0, 0, 0)$.
$A$ plane $Ax + By + Cz + D = 0$ contains the $x$-axis if and only if the coefficients of $y$ and $z$ are such that the equation is satisfied for any point $(x, 0, 0)$ on the $x$-axis.
Substituting $(x, 0, 0)$ into $3y + 4z = 0$,we get $3(0) + 4(0) = 0$,which is $0 = 0$.
This holds true for all values of $x$.
Therefore,the plane contains the $x$-axis.

(B) Let the coordinates of point $P$ be $(x, y, z)$.
Given $PA^2 - PB^2 = 6c$.
Using the distance formula,$PA^2 = (x - 1)^2 + (y - 3)^2 + (z - 5)^2$ and $PB^2 = (x + 2)^2 + (y - 3)^2 + (z + 4)^2$.
Substituting these into the equation:
$[(x - 1)^2 + (y - 3)^2 + (z - 5)^2] - [(x + 2)^2 + (y - 3)^2 + (z + 4)^2] = 6c$.
The $(y - 3)^2$ terms cancel out:
$(x - 1)^2 - (x + 2)^2 + (z - 5)^2 - (z + 4)^2 = 6c$.
Expanding the squares:
$(x^2 - 2x + 1) - (x^2 + 4x + 4) + (z^2 - 10z + 25) - (z^2 + 8z + 16) = 6c$.
Simplifying the expression:
$-6x - 3 - 18z + 9 = 6c$.
$-6x - 18z + 6 = 6c$.
Dividing by $-6$:
$x + 3z - 1 = -c$,which rearranges to $x + 3z - 1 + c = 0$.

(D) The normal vectors of the planes are $\vec{n_1} = (1, -1, 1)$,$\vec{n_2} = (1, 1, -1)$,and $\vec{n_3} = (1, -3, 3)$.
The direction vectors of the lines of intersection are given by the cross products of the normals:
$\vec{v_1} = \vec{n_2} \times \vec{n_3} = \begin{vmatrix} i & j & k \\ 1 & 1 & -1 \\ 1 & -3 & 3 \end{vmatrix} = (0, -4, -4) \parallel (0, 1, 1)$.
$\vec{v_2} = \vec{n_3} \times \vec{n_1} = \begin{vmatrix} i & j & k \\ 1 & -3 & 3 \\ 1 & -1 & 1 \end{vmatrix} = (0, 2, 2) \parallel (0, 1, 1)$.
$\vec{v_3} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} i & j & k \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = (0, 2, 2) \parallel (0, 1, 1)$.
Since all direction vectors are parallel to $(0, 1, 1)$,the lines $L_1, L_2, L_3$ are all parallel to each other. Thus,Statement-$1$ is false.
For Statement-$2$,we check the determinant of the coefficient matrix:
$D = \begin{vmatrix} 1 & -1 & 1 \\ 1 & 1 & -1 \\ 1 & -3 & 3 \end{vmatrix} = 1(3-3) + 1(3+1) + 1(-3-1) = 0 + 4 - 4 = 0$.
Since $D=0$,the planes do not have a unique common point. Checking for consistency: $P_1+P_2 = 2x = 0 \implies x=0$. Substituting $x=0$ into $P_1$ and $P_2$: $-y+z=1$ and $y-z=-1$. These are the same equation. Substituting $x=0$ into $P_3$: $-3y+3z=2 \implies y-z = -2/3$. Since $1 \neq -2/3$,there is no common point. Thus,Statement-$2$ is true.

(B) The equation of a plane passing through the point $(2, 2, 1)$ is given by $a(x - 2) + b(y - 2) + c(z - 1) = 0$.
Since the plane passes through $(9, 3, 6)$,we have $a(9 - 2) + b(3 - 2) + c(6 - 1) = 0$,which simplifies to $7a + b + 5c = 0$.
Since the plane is perpendicular to $2x + 6y + 6z - 1 = 0$,the normal vectors are perpendicular,so $2a + 6b + 6c = 0$,which simplifies to $a + 3b + 3c = 0$.
Using the cross product of the direction vectors $(7, 1, 5)$ and $(1, 3, 3)$ to find the normal vector $(a, b, c)$:
$a = (1 \times 3) - (5 \times 3) = 3 - 15 = -12$
$b = (5 \times 1) - (7 \times 3) = 5 - 21 = -16$
$c = (7 \times 3) - (1 \times 1) = 21 - 1 = 20$
Dividing by $-4$,we get the direction ratios $(a, b, c) = (3, 4, -5)$.
Substituting these into the plane equation: $3(x - 2) + 4(y - 2) - 5(z - 1) = 0$.
$3x - 6 + 4y - 8 - 5z + 5 = 0 \Rightarrow 3x + 4y - 5z = 9$.

(B) The equation of a plane parallel to $x - 2y + 2z - 5 = 0$ is of the form $x - 2y + 2z + k = 0$.
Given that the distance from the origin $(0, 0, 0)$ to this plane is $1$ unit.
The formula for the distance from a point $(x_1, y_1, z_1)$ to a plane $Ax + By + Cz + D = 0$ is $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.
Substituting the values,we get $1 = \frac{|1(0) - 2(0) + 2(0) + k|}{\sqrt{1^2 + (-2)^2 + 2^2}}$.
$1 = \frac{|k|}{\sqrt{1 + 4 + 4}} = \frac{|k|}{\sqrt{9}} = \frac{|k|}{3}$.
Therefore,$|k| = 3$,which implies $k = 3$ or $k = -3$.
Substituting these values back into the equation,we get $x - 2y + 2z + 3 = 0$ or $x - 2y + 2z - 3 = 0$.
Comparing with the given options,$x - 2y + 2z - 3 = 0$ is the correct choice.

(C) The angle $\theta$ between two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ is given by the formula:
$\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$
For the given planes $x + y + 2z = 9$ and $2x - y + z = 15$,we have:
$a_1 = 1, b_1 = 1, c_1 = 2$
$a_2 = 2, b_2 = -1, c_2 = 1$
Substituting these values into the formula:
$\cos \theta = \frac{|(1)(2) + (1)(-1) + (2)(1)|}{\sqrt{1^2 + 1^2 + 2^2} \sqrt{2^2 + (-1)^2 + 1^2}}$
$\cos \theta = \frac{|2 - 1 + 2|}{\sqrt{1 + 1 + 4} \sqrt{4 + 1 + 1}}$
$\cos \theta = \frac{3}{\sqrt{6} \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$
Since $\cos \theta = \frac{1}{2}$,the angle $\theta = \frac{\pi}{3}$.

(B) The given planes are $P_1: 2x - y + z - 2 = 0$ and $P_2: x + 2y - z - 3 = 0$.
The equation of the angle bisectors is given by $\frac{2x - y + z - 2}{\sqrt{2^2 + (-1)^2 + 1^2}} = \pm \frac{x + 2y - z - 3}{\sqrt{1^2 + 2^2 + (-1)^2}}$.
Since the denominators are equal,we have $2x - y + z - 2 = \pm (x + 2y - z - 3)$.
Case $1$ (Positive sign): $2x - y + z - 2 = x + 2y - z - 3 \implies x - 3y + 2z + 1 = 0$.
Case $2$ (Negative sign): $2x - y + z - 2 = -x - 2y + z + 3 \implies 3x + y - 5 = 0$.
To identify the acute angle bisector,check the sign of $a_1a_2 + b_1b_2 + c_1c_2$ where $a_1, b_1, c_1$ and $a_2, b_2, c_2$ are coefficients of $x, y, z$ after making the constant terms positive.
Rewriting planes as $P_1: -2x + y - z + 2 = 0$ and $P_2: x + 2y - z - 3 = 0$ is not standard; let's use $P_1: -2x + y - z + 2 = 0$ and $P_2: x + 2y - z - 3 = 0$.
Actually,for $P_1: 2x - y + z - 2 = 0$ and $P_2: x + 2y - z - 3 = 0$,$a_1a_2 + b_1b_2 + c_1c_2 = (2)(1) + (-1)(2) + (1)(-1) = 2 - 2 - 1 = -1 < 0$.
Since the product is negative,the positive sign gives the obtuse angle bisector and the negative sign gives the acute angle bisector.
Thus,the acute angle bisector is $3x + y - 5 = 0$.

(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 $p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}$,which implies $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{p^2}$.
Since the planes are drawn parallel to the coordinate planes through $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$,the point of intersection of these planes is $(a, b, c)$.
Replacing $(a, b, c)$ with $(x, y, z)$,the locus is $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{1}{p^2}$.

(D) The equation of any plane parallel to $3x - 5y + 2z = 11$ is of the form $3x - 5y + 2z = k$.
Since this plane passes through the point $(1, 2, -3)$,we substitute these coordinates into the equation:
$3(1) - 5(2) + 2(-3) = k$
$3 - 10 - 6 = k$
$k = -13$
Substituting $k = -13$ back into the equation,we get $3x - 5y + 2z = -13$,which can be rewritten as $3x - 5y + 2z + 13 = 0$.

(B) 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 $x, y,$ and $z$ axes respectively.
Given that the plane makes equal intercepts of unit length,we have $a = 1, b = 1, c = 1$.
Substituting these values into the intercept form equation:
$\frac{x}{1} + \frac{y}{1} + \frac{z}{1} = 1$
Therefore,the equation of the plane is $x + y + z = 1$.

(B) The given equation of the plane is $\bar{r} \cdot (2, -3, 4) = 12$.
Substituting $\bar{r} = (x, y, z)$,we get $2x - 3y + 4z = 12$.
To find the intercepts,we write the equation in the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Dividing the equation $2x - 3y + 4z = 12$ by $12$,we get $\frac{2x}{12} - \frac{3y}{12} + \frac{4z}{12} = 1$.
This simplifies to $\frac{x}{6} + \frac{y}{-4} + \frac{z}{3} = 1$.
Comparing this with the standard intercept form,the intercepts are $a = 6$,$b = -4$,and $c = 3$.

(C) The given equation of the plane is $\vec{r} = (1 + \lambda - \mu)\hat{i} + (2 - \lambda)\hat{j} + (3 - 2\lambda + 2\mu)\hat{k}$.
We can rewrite this as $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} - 2\hat{k}) + \mu(-\hat{i} + 2\hat{k})$.
This represents a plane passing through the point $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and parallel to the vectors $\vec{b} = \hat{i} - \hat{j} - 2\hat{k}$ 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 vector 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 $2x + z = 5$.

(A) The line passes through the origin and makes equal angles with the axes,so its direction ratios are $(1, 1, 1)$.
The equation of the line is $\frac{x}{1} = \frac{y}{1} = \frac{z}{1}$.
The point $P$ is $(a, a, a)$.
The plane passes through $P(a, a, a)$ and is perpendicular to $OP$. The direction ratios of the normal to the plane are the same as the direction ratios of the line $OP$,which are $(1, 1, 1)$.
The equation of the plane is $1(x - a) + 1(y - a) + 1(z - a) = 0$.
Simplifying this,we get $x + y + z = 3a$.
Dividing by $3a$,we get $\frac{x}{3a} + \frac{y}{3a} + \frac{z}{3a} = 1$.
The intercepts on the axes are $3a, 3a, 3a$.
The sum of the reciprocals of the intercepts is $\frac{1}{3a} + \frac{1}{3a} + \frac{1}{3a} = \frac{3}{3a} = \frac{1}{a}$.

(D) Let the equation of the variable plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
This plane meets the coordinate axes at $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.
Let $(\alpha, \beta, \gamma)$ be the coordinates of the centroid of $\Delta ABC$. Then,
$\alpha = \frac{a}{3}, \beta = \frac{b}{3}, \gamma = \frac{c}{3} \implies a = 3\alpha, b = 3\beta, c = 3\gamma \dots (i)$
The distance of the plane from the origin is $k$.
$\therefore \left| \frac{\frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} \right| = k$
$\implies \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{k^2}$
Substituting values from $(i)$:
$\frac{1}{(3\alpha)^2} + \frac{1}{(3\beta)^2} + \frac{1}{(3\gamma)^2} = \frac{1}{k^2}$
$\frac{1}{9\alpha^2} + \frac{1}{9\beta^2} + \frac{1}{9\gamma^2} = \frac{1}{k^2}$
$\alpha^{-2} + \beta^{-2} + \gamma^{-2} = 9k^{-2}$
Thus,the locus of $(\alpha, \beta, \gamma)$ is $x^{-2} + y^{-2} + z^{-2} = 9k^{-2}$.

(D) The general equation of a plane parallel to $3x - 4y + 5z - 6 = 0$ is given by $3x - 4y + 5z + \lambda = 0$.
Since the plane passes through the origin $(0, 0, 0)$,we substitute these coordinates into the equation:
$3(0) - 4(0) + 5(0) + \lambda = 0$
This gives $\lambda = 0$.
Substituting $\lambda = 0$ back into the general equation,we get the required plane equation: $3x - 4y + 5z = 0$.

(D) The equation of the plane bisecting the angle between the planes $P_1: 2x - y + 2z + 3 = 0$ and $P_2: 3x - 2y + 6z + 8 = 0$ is given by the formula:
$\frac{2x - y + 2z + 3}{\sqrt{2^2 + (-1)^2 + 2^2}} = \pm \frac{3x - 2y + 6z + 8}{\sqrt{3^2 + (-2)^2 + 6^2}}$
Calculating the denominators:
$\sqrt{4 + 1 + 4} = \sqrt{9} = 3$
$\sqrt{9 + 4 + 36} = \sqrt{49} = 7$
So,the equation becomes:
$\frac{2x - y + 2z + 3}{3} = \pm \frac{3x - 2y + 6z + 8}{7}$
Cross-multiplying:
$7(2x - y + 2z + 3) = \pm 3(3x - 2y + 6z + 8)$
Case $1$ (Positive sign):
$14x - 7y + 14z + 21 = 9x - 6y + 18z + 24$
$5x - y - 4z - 3 = 0$
Case $2$ (Negative sign):
$14x - 7y + 14z + 21 = -(9x - 6y + 18z + 24)$
$14x - 7y + 14z + 21 = -9x + 6y - 18z - 24$
$23x - 13y + 32z + 45 = 0$
Comparing with the given options,the correct equation is $23x - 13y + 32z + 45 = 0$.

(B) Three planes $a_1x + b_1y + c_1z = 0$,$a_2x + b_2y + c_2z = 0$,and $a_3x + b_3y + c_3z = 0$ intersect in a single line if the determinant of their coefficients is zero,i.e.,$\begin{vmatrix} k & 4 & 1 \\ 4 & k & 2 \\ 2 & 2 & 1 \end{vmatrix} = 0$.
Expanding the determinant along the first row:
$k(k(1) - 2(2)) - 4(4(1) - 2(2)) + 1(4(2) - 2(k)) = 0$
$k(k - 4) - 4(4 - 4) + 1(8 - 2k) = 0$
$k^2 - 4k - 0 + 8 - 2k = 0$
$k^2 - 6k + 8 = 0$
Factoring the quadratic equation:
$(k - 2)(k - 4) = 0$
So,$k = 2$ or $k = 4$.
Checking the options,$k = 2$ is provided as option $B$.

(A) Let the normal vector of the required plane be $\vec{n} = (a, b, c)$.
Since the plane is perpendicular to the planes with normal vectors $\vec{n_1} = (2, -2, 1)$ and $\vec{n_2} = (1, -1, 2)$,the vector $\vec{n}$ is parallel to $\vec{n_1} \times \vec{n_2}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & -1 & 2 \end{vmatrix} = \hat{i}(-4 + 1) - \hat{j}(4 - 1) + \hat{k}(-2 + 2) = -3\hat{i} - 3\hat{j} + 0\hat{k}$.
We can take the normal vector as $\vec{n} = (1, 1, 0)$.
The equation of the plane passing through $(1, -2, 1)$ is $1(x - 1) + 1(y + 2) + 0(z - 1) = 0$,which simplifies to $x + y + 1 = 0$.
The distance of the plane $x + y + 1 = 0$ from the point $(1, 2, 2)$ is given by $d = \frac{|1(1) + 1(2) + 1|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.

(A) Let the equation of the plane be $f(x, y, z) = 2x + 2y - 2z - 1 = 0$.
To check if two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ lie on opposite sides of a plane,we evaluate $f(x_1, y_1, z_1)$ and $f(x_2, y_2, z_2)$. If the results have opposite signs,the points lie on opposite sides.
For point $(2, 1, 5)$: $f(2, 1, 5) = 2(2) + 2(1) - 2(5) - 1 = 4 + 2 - 10 - 1 = -5$.
For point $(3, 4, 3)$: $f(3, 4, 3) = 2(3) + 2(4) - 2(3) - 1 = 6 + 8 - 6 - 1 = 7$.
Since $f(2, 1, 5) = -5$ and $f(3, 4, 3) = 7$ have opposite signs,the points lie on opposite sides of the plane.
The algebraic perpendicular distance is given by $\frac{f(x, y, z)}{\sqrt{a^2 + b^2 + c^2}}$. Since the denominator is always positive,the sign of the distance depends solely on the sign of $f(x, y, z)$.
Thus,both the Assertion and the Reason are true,and the Reason correctly explains the Assertion.

(B) The equation of the plane passing through the intersection of the planes $\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 5$ and $\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3$ is given by:
$[\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) - 5] + \lambda [\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) - 3] = 0$
$\Rightarrow \vec{r} \cdot [(1 + 2\lambda)\hat{i} + (3 - \lambda)\hat{j} + (-1 + \lambda)\hat{k}] = 5 + 3\lambda \quad (i)$
Since the plane passes through the point $(2, 1, -2)$,the position vector is $\vec{r} = 2\hat{i} + \hat{j} - 2\hat{k}$. Substituting this into equation $(i)$:
$(2\hat{i} + \hat{j} - 2\hat{k}) \cdot [(1 + 2\lambda)\hat{i} + (3 - \lambda)\hat{j} + (-1 + \lambda)\hat{k}] = 5 + 3\lambda$
$2(1 + 2\lambda) + 1(3 - \lambda) - 2(-1 + \lambda) = 5 + 3\lambda$
$2 + 4\lambda + 3 - \lambda + 2 - 2\lambda = 5 + 3\lambda$
$7 + \lambda = 5 + 3\lambda$
$2 = 2\lambda \Rightarrow \lambda = 1$
Substituting $\lambda = 1$ in equation $(i)$:
$\vec{r} \cdot [(1 + 2(1))\hat{i} + (3 - 1)\hat{j} + (-1 + 1)\hat{k}] = 5 + 3(1)$
$\vec{r} \cdot (3\hat{i} + 2\hat{j} + 0\hat{k}) = 8$.

(D) The equations of the planes are $\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19 = 0$ and $\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3 = 0$.
The equation of the angle bisector planes is given by:
$\frac{\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19}{\sqrt{1^2 + 2^2 + 2^2}} = \pm \frac{\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3}{\sqrt{4^2 + (-3)^2 + 12^2}}$
$\frac{\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19}{3} = \pm \frac{\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3}{13}$
Case $1$ (Positive sign):
$13(\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19) = 3(\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3)$
$\vec{r} \cdot (13\hat{i} + 26\hat{j} + 26\hat{k} - 12\hat{i} + 9\hat{j} - 36\hat{k}) = 9 + 247$
$\vec{r} \cdot (\hat{i} + 35\hat{j} - 10\hat{k}) = 256$
Case $2$ (Negative sign):
$13(\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19) = -3(\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3)$
$\vec{r} \cdot (13\hat{i} + 26\hat{j} + 26\hat{k} + 12\hat{i} - 9\hat{j} + 36\hat{k}) = -9 + 247$
$\vec{r} \cdot (25\hat{i} + 17\hat{j} + 62\hat{k}) = 238$
Comparing with the options,the correct equation is $\vec{r} \cdot (25\hat{i} + 17\hat{j} + 62\hat{k}) = 238$.

(A) Let the points be $A(\hat{i} + \hat{j} - 2\hat{k})$,$B(2\hat{i} - \hat{j} + \hat{k})$,and $C(\hat{i} + 2\hat{j} + \hat{k})$.
The vectors lying in the plane are:
$\vec{AB} = (2\hat{i} - \hat{j} + \hat{k}) - (\hat{i} + \hat{j} - 2\hat{k}) = \hat{i} - 2\hat{j} + 3\hat{k}$
$\vec{AC} = (\hat{i} + 2\hat{j} + \hat{k}) - (\hat{i} + \hat{j} - 2\hat{k}) = 0\hat{i} + \hat{j} + 3\hat{k}$
The normal vector $\vec{n}$ to the plane is given by the cross product $\vec{AB} \times \vec{AC}$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 0 & 1 & 3 \end{vmatrix} = \hat{i}(-6 - 3) - \hat{j}(3 - 0) + \hat{k}(1 - 0) = -9\hat{i} - 3\hat{j} + \hat{k}$.
The vector equation of the plane passing through point $\vec{a} = \hat{i} + \hat{j} - 2\hat{k}$ with normal $\vec{n} = -9\hat{i} - 3\hat{j} + \hat{k}$ is $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$,which implies $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
Calculating $\vec{a} \cdot \vec{n} = (\hat{i} + \hat{j} - 2\hat{k}) \cdot (-9\hat{i} - 3\hat{j} + \hat{k}) = -9 - 3 - 2 = -14$.
Thus,$\vec{r} \cdot (-9\hat{i} - 3\hat{j} + \hat{k}) = -14$,or $\vec{r} \cdot (9\hat{i} + 3\hat{j} - \hat{k}) = 14$.

(D) To check Statement $-1$,we find the mirror image of point $B(1,3,4)$ in the plane $x-y+z-5=0.$
Using the formula $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = -2 \frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$,we get:
$\frac{x-1}{1} = \frac{y-3}{-1} = \frac{z-4}{1} = -2 \frac{1-3+4-5}{1^2+(-1)^2+1^2} = -2 \frac{-3}{3} = 2.$
Thus,$x-1=2 \Rightarrow x=3$,$y-3=-2 \Rightarrow y=1$,$z-4=2 \Rightarrow z=6$.
The image is $(3,1,6)$,which is point $A$. So,Statement $-1$ is true.
For Statement $-2$,the midpoint of $AB$ is $(\frac{3+1}{2}, \frac{1+3}{2}, \frac{6+4}{2}) = (2,2,5)$.
Checking if this point lies on the plane: $2-2+5 = 5$. Since it satisfies the equation,the plane bisects the segment $AB$. So,Statement $-2$ is true.
Since the mirror image definition requires the line segment $AB$ to be perpendicular to the plane $AND$ the midpoint to lie on the plane,Statement $-2$ (which only confirms the midpoint) is a necessary condition but not the complete definition of a mirror image. However,in the context of these types of questions,Statement $-2$ provides the geometric basis for the bisection property,which is part of the mirror image condition. Thus,Statement $-2$ is a correct explanation.

(A) Let the equation of the plane parallel to $x - 2y + 2z - 5 = 0$ be $x - 2y + 2z + k = 0 \dots (i)$
The perpendicular distance from the origin $(0, 0, 0)$ to the plane $(i)$ is given by the formula $d = \frac{|k|}{\sqrt{a^2 + b^2 + c^2}}$.
Given that the distance is $1$,we have:
$\frac{|k|}{\sqrt{1^2 + (-2)^2 + 2^2}} = 1$
$\frac{|k|}{\sqrt{1 + 4 + 4}} = 1$
$\frac{|k|}{\sqrt{9}} = 1$
$\frac{|k|}{3} = 1$
$|k| = 3$
Therefore,$k = 3$ or $k = -3$.
Substituting these values into equation $(i)$,we get the possible equations of the plane as $x - 2y + 2z + 3 = 0$ or $x - 2y + 2z - 3 = 0$.
Comparing this with the given options,the correct equation is $x - 2y + 2z - 3 = 0$.

(A) The normal vector $\vec{n}$ to the plane $AOB$ is given by the cross product of the vectors $\vec{OA}$ and $\vec{OB}$.
Given $\vec{OA} = (1, -2, -1)$ and $\vec{OB} = (3, -2, 3)$.
$\vec{n} = \vec{OA} \times \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -1 \\ 3 & -2 & 3 \end{vmatrix} = \hat{i}(-6 - 2) - \hat{j}(3 + 3) + \hat{k}(-2 + 6) = -8\hat{i} - 6\hat{j} + 4\hat{k}$.
Dividing by $-2$ to simplify,the normal vector is proportional to $(4, 3, -2)$.
The magnitude of the vector $(4, 3, -2)$ is $\sqrt{4^2 + 3^2 + (-2)^2} = \sqrt{16 + 9 + 4} = \sqrt{29}$.
The direction cosines are $\left( \frac{4}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{-2}{\sqrt{29}} \right)$.

(D) The given equations are $x + y + z + 2 = 0$ and $x + y + z + 3 = 0$.
These equations represent two planes in $3D$ space.
The normal vectors to these planes are $\vec{n_1} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{n_2} = \hat{i} + \hat{j} + \hat{k}$.
Since the normal vectors are identical,the planes are parallel to each other.
Two distinct parallel planes do not intersect at any point in space.
Therefore,the system of these two equations represents an empty set,meaning they represent nothing in space.

(B) 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 $(0, 0, 0)$,the perpendicular distance $d$ is given by $d = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 1$.
Squaring both sides,we get $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = 1$ ... $(i)$.
The coordinates of the points $P, Q,$ and $R$ are $(a, 0, 0), (0, b, 0),$ and $(0, 0, c)$ respectively.
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,$a = 3x, b = 3y, c = 3z$.
Substituting these into equation $(i)$,we get $\frac{1}{(3x)^2} + \frac{1}{(3y)^2} + \frac{1}{(3z)^2} = 1$.
This simplifies to $\frac{1}{9x^2} + \frac{1}{9y^2} + \frac{1}{9z^2} = 1$,which implies $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 9$.
Comparing this with the given locus $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = k$,we find $k = 9$.

(C) The equation of the plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b,$ and $c$ are the intercepts on the $x, y,$ and $z$ axes respectively.
Comparing this with the given equation $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$,we get $a = 2, b = 3,$ and $c = 4$.
The coordinates of the points where the plane cuts the axes are $A(2, 0, 0), B(0, 3, 0),$ and $C(0, 0, 4)$.
The area of a triangle with vertices $(a, 0, 0), (0, b, 0),$ and $(0, 0, c)$ is given by the formula $\text{Area} = \frac{1}{2} \sqrt{a^2b^2 + b^2c^2 + c^2a^2}$.
Substituting the values $a=2, b=3, c=4$:
$\text{Area} = \frac{1}{2} \sqrt{(2^2 \times 3^2) + (3^2 \times 4^2) + (4^2 \times 2^2)}$
$\text{Area} = \frac{1}{2} \sqrt{(4 \times 9) + (9 \times 16) + (16 \times 4)}$
$\text{Area} = \frac{1}{2} \sqrt{36 + 144 + 64}$
$\text{Area} = \frac{1}{2} \sqrt{244}$
$\text{Area} = \frac{1}{2} \sqrt{4 \times 61} = \frac{1}{2} \times 2 \sqrt{61} = \sqrt{61}$.

(C) Given the equation $f(x, y, z) = xy + xz = 0$.
Factoring the expression,we get $x(y + z) = 0$.
This equation is satisfied if $x = 0$ or $y + z = 0$.
In three-dimensional space,$x = 0$ represents the $yz$-plane.
The equation $y + z = 0$ represents a plane passing through the $x$-axis.
The normal vector to the plane $x = 0$ is $\vec{n_1} = (1, 0, 0)$.
The normal vector to the plane $y + z = 0$ is $\vec{n_2} = (0, 1, 1)$.
The dot product of the normal vectors is $\vec{n_1} \cdot \vec{n_2} = (1)(0) + (0)(1) + (0)(1) = 0$.
Since the dot product of the normal vectors is $0$,the two planes are perpendicular to each other.

(B) Let $\gamma$ be the angle made by $n$ with the $z$-axis. The direction cosines of $n$ are $l = \cos 45^\circ = \frac{1}{\sqrt{2}}$,$m = \cos 60^\circ = \frac{1}{2}$,and $n = \cos \gamma$.
Since $l^2 + m^2 + n^2 = 1$,we have $(\frac{1}{\sqrt{2}})^2 + (\frac{1}{2})^2 + n^2 = 1$.
$\Rightarrow \frac{1}{2} + \frac{1}{4} + n^2 = 1 \Rightarrow n^2 = 1 - \frac{3}{4} = \frac{1}{4}$.
Since $\gamma$ is acute,$n = \cos \gamma = \frac{1}{2}$.
Given $|n| = 8$,the vector $n = |n|(l i + m j + n k) = 8(\frac{1}{\sqrt{2}} i + \frac{1}{2} j + \frac{1}{2} k) = 4\sqrt{2} i + 4 j + 4 k$.
The plane passes through point $A$ with position vector $a = \sqrt{2} i - j + k$ and is normal to $n$.
The vector equation of the plane is $r \cdot n = a \cdot n$.
$r \cdot (4\sqrt{2} i + 4 j + 4 k) = (\sqrt{2} i - j + k) \cdot (4\sqrt{2} i + 4 j + 4 k)$.
$r \cdot (4\sqrt{2} i + 4 j + 4 k) = (\sqrt{2})(4\sqrt{2}) + (-1)(4) + (1)(4) = 8 - 4 + 4 = 8$.
Dividing by $4$,we get $r \cdot (\sqrt{2} i + j + k) = 2$.

(A) Let the normal vector to the plane be $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}.$ The equation of the plane passing through $A(2, 1, -3)$ is $a(x-2) + b(y-1) + c(z+3) = 0,$ which simplifies to $ax + by + cz = 2a + b - 3c.$
The distance $d$ of this plane from the origin $(0, 0, 0)$ is given by $d = \frac{|2a + b - 3c|}{\sqrt{a^2 + b^2 + c^2}}.$
By the Cauchy-Schwarz inequality,for any vectors $\vec{u} = (2, 1, -3)$ and $\vec{n} = (a, b, c),$ we have $|\vec{u} \cdot \vec{n}| \leq |\vec{u}| |\vec{n}|.$ Thus,$d = \frac{|\vec{u} \cdot \vec{n}|}{|\vec{n}|} \leq |\vec{u}| = \sqrt{2^2 + 1^2 + (-3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14}.$
The distance is maximum when the normal vector $\vec{n}$ is parallel to the position vector $\vec{OA} = 2\hat{i} + \hat{j} - 3\hat{k}.$
Setting $\vec{n} = (2, 1, -3),$ the equation of the plane becomes $2(x-2) + 1(y-1) - 3(z+3) = 0,$
$2x - 4 + y - 1 - 3z - 9 = 0,$
$2x + y - 3z = 14.$

(A) Let the vertex $M$ be at the origin $(0, 0, 0)$. Let the lengths of the edges $ML, MN, MO$ be $a, b, c$ respectively.
The equation of the plane passing through $L(a, 0, 0), N(0, b, 0)$ and $O(0, 0, c)$ is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
The altitude from $O$ to face $MLN$ is $c = 1$. The altitude from $L$ to face $MNO$ is $a = 2$. The altitude from $N$ to face $MLO$ is $b = 3$.
Thus,the equation of the face $LNO$ is $\frac{x}{2} + \frac{y}{3} + \frac{z}{1} = 1$,which simplifies to $\frac{x}{2} + \frac{y}{3} + z - 1 = 0$.
The length of the altitude $h$ from $M(0, 0, 0)$ to the face $LNO$ is given by the distance formula:
$h = \frac{|-1|}{\sqrt{(\frac{1}{2})^2 + (\frac{1}{3})^2 + (1)^2}} = \frac{1}{\sqrt{\frac{1}{4} + \frac{1}{9} + 1}} = \frac{1}{\sqrt{\frac{9+4+36}{36}}} = \frac{1}{\sqrt{\frac{49}{36}}} = \frac{6}{7} \text{ units}$.

(A) Let the point be $A(2, 1, -3)$ and the origin be $O(0, 0, 0)$.
For a plane passing through a fixed point $A$ to have a maximum distance from the origin $O$,the normal vector $\vec{n}$ of the plane must be the vector $\vec{OA}$.
Thus,the normal vector is $\vec{n} = \vec{OA} = 2\hat{i} + \hat{j} - 3\hat{k}$.
The equation of a plane passing through point $\vec{a}$ with normal vector $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$,which is $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
Here,$\vec{a} = 2\hat{i} + \hat{j} - 3\hat{k}$ and $\vec{n} = 2\hat{i} + \hat{j} - 3\hat{k}$.
So,the equation is $2x + y - 3z = (2)(2) + (1)(1) + (-3)(-3)$.
$2x + y - 3z = 4 + 1 + 9 = 14$.
Therefore,the equation of the plane is $2x + y - 3z = 14$.

(C) The given planes are:
$y + z = 0$ ... $(1)$
$z + x = 0$ ... $(2)$
$x + y = 0$ ... $(3)$
$x + y + z = 2$ ... $(4)$
To find the vertices of the tetrahedron,we solve the systems of equations formed by taking three planes at a time:
- Intersection of $(1), (2), (3)$: $x+y=0, y+z=0, z+x=0 \implies x=y=z=0$. Vertex $V_1 = (0, 0, 0)$.
- Intersection of $(1), (2), (4)$: $y+z=0, z+x=0, x+y+z=2$. Substituting $y=-z$ and $x=-z$ into $(4)$,we get $-z-z+z=2 \implies z=-2$. Thus,$x=2, y=2$. Vertex $V_2 = (2, 2, -2)$.
- Intersection of $(1), (3), (4)$: $y+z=0, x+y=0, x+y+z=2$. Substituting $z=-y$ and $x=-y$ into $(4)$,we get $-y-y+y=2 \implies y=-2$. Thus,$x=2, z=2$. Vertex $V_3 = (2, -2, 2)$.
- Intersection of $(2), (3), (4)$: $z+x=0, x+y=0, x+y+z=2$. Substituting $z=-x$ and $y=-x$ into $(4)$,we get $x-x-x=2 \implies x=-2$. Thus,$y=2, z=2$. Vertex $V_4 = (-2, 2, 2)$.
The volume of the tetrahedron with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3), (x_4, y_4, z_4)$ is given by $\frac{1}{6} |det(V_2-V_1, V_3-V_1, V_4-V_1)|$.
Volume $= \frac{1}{6} \left| \begin{matrix} 2 & 2 & -2 \\ 2 & -2 & 2 \\ -2 & 2 & 2 \end{matrix} \right| = \frac{1}{6} |2(-4-4) - 2(4+4) - 2(4-4)| = \frac{1}{6} |-16 - 16 - 0| = \frac{32}{6} = \frac{16}{3}$.

(A) The general equation of a pair of planes passing through the origin is $ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy = 0$.
Comparing this with the given equation $2x^2 - 6y^2 - 12z^2 + 18yz + 2zx + xy = 0$,we have $a=2, b=-6, c=-12, f=9, g=1, h=0.5$.
The planes are represented by $2x^2 + xy - 6y^2 + 2zx + 18yz - 12z^2 = 0$.
Factoring the quadratic expression,we get $(x + 2y - 2z)(2x - 3y + 6z) = 0$.
Thus,the equations of the planes are $P_1: x + 2y - 2z = 0$ and $P_2: 2x - 3y + 6z = 0$.
The normal vectors to these planes are $\vec{n_1} = (1, 2, -2)$ and $\vec{n_2} = (2, -3, 6)$.
The cosine of the angle $\alpha$ between the planes is given by $\cos \alpha = \left| \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}| |\vec{n_2}|} \right|$.
$\vec{n_1} \cdot \vec{n_2} = (1)(2) + (2)(-3) + (-2)(6) = 2 - 6 - 12 = -16$.
$|\vec{n_1}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = 3$.
$|\vec{n_2}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = 7$.
Therefore,$\cos \alpha = \left| \frac{-16}{3 \times 7} \right| = \frac{16}{21}$.

(B) Let the equation of the plane be $a(x+1) + b(y-3) + c(z+2) = 0$.
Since the plane contains the line,the normal vector is perpendicular to the direction vector $(-3, 2, 1)$,so $-3a + 2b + c = 0$.
The plane passes through $(0, 7, -7)$,so $a(0+1) + b(7-3) + c(-7+2) = 0$,which gives $a + 4b - 5c = 0$.
Solving the system $-3a + 2b + c = 0$ and $a + 4b - 5c = 0$ using cross product,we get the normal vector $\vec{n} = (1, 1, 1)$.
The equation of the plane is $1(x+1) + 1(y-3) + 1(z+2) = 0$,which simplifies to $x + y + z = 0$.
The line passing through $(1, 2, -1)$ and perpendicular to the plane is $\frac{x-1}{1} = \frac{y-2}{1} = \frac{z+1}{1} = \lambda$.
Any point on this line is $(\lambda+1, \lambda+2, \lambda-1)$.
For the foot of the perpendicular,this point lies on the plane: $(\lambda+1) + (\lambda+2) + (\lambda-1) = 0$,so $3\lambda + 2 = 0$,which gives $\lambda = \frac{-2}{3}$.
The foot of the perpendicular is $(\frac{1}{3}, \frac{4}{3}, \frac{-5}{3})$.
Let the image be $(x', y', z')$. The midpoint of the point and its image is the foot of the perpendicular: $\frac{x'+1}{2} = \frac{1}{3} \Rightarrow x' = \frac{-1}{3}$,$\frac{y'+2}{2} = \frac{4}{3} \Rightarrow y' = \frac{2}{3}$,and $\frac{z'-1}{2} = \frac{-5}{3} \Rightarrow z' = \frac{-7}{3}$.
Thus,the image is $\left( \frac{-1}{3}, \frac{2}{3}, \frac{-7}{3} \right)$.

(B) The coordinates of point $P$ are $(a, b, c)$.
Since $PA$ is perpendicular to the $yz$-plane,the coordinates of $A$ are $(0, b, c)$.
Since $PB$ is perpendicular to the $zx$-plane,the coordinates of $B$ are $(a, 0, c)$.
The origin $O$ is $(0, 0, 0)$.
The equation of a plane passing through the origin is $px + qy + rz = 0$.
Since the plane passes through $A(0, b, c)$,we have $p(0) + q(b) + r(c) = 0$,which implies $qb + rc = 0$.
Since the plane passes through $B(a, 0, c)$,we have $p(a) + q(0) + r(c) = 0$,which implies $pa + rc = 0$.
From these equations,$qb = -rc$ and $pa = -rc$.
Thus,$pa = qb = -rc = k$ (where $k$ is a constant).
Then $p = k/a$,$q = k/b$,and $r = -k/c$.
Substituting these into the plane equation: $(k/a)x + (k/b)y - (k/c)z = 0$.
Dividing by $k$ and multiplying by $abc$,we get $bcx + acy - abz = 0$.

(B) Let the set of points on $P_1$ be $S_1 = \{A, B, C\}$ and the set of points on $P_2$ be $S_2 = \{D, E, F\}$.
$A$ tetrahedron is formed by selecting $4$ non-coplanar points.
Since $3$ points on $P_1$ are coplanar and $3$ points on $P_2$ are coplanar,we cannot select $4$ points from the same plane.
Thus,we must select points from both planes.
The possible combinations are:
$1$. Select $3$ points from $P_1$ and $1$ point from $P_2$: ${}^3C_3 \times {}^3C_1 = 1 \times 3 = 3$.
$2$. Select $2$ points from $P_1$ and $2$ points from $P_2$: ${}^3C_2 \times {}^3C_2 = 3 \times 3 = 9$.
$3$. Select $1$ point from $P_1$ and $3$ points from $P_2$: ${}^3C_1 \times {}^3C_3 = 3 \times 1 = 3$.
Total number of tetrahedrons $= 3 + 9 + 3 = 15$.