Plane Questions in English

(A) Given points are $A(-1, 2, 3)$,$B(1, 1, 1)$,and $C(2, -1, 3)$.
First,we find two vectors lying on the plane:
$\overrightarrow{AB} = (1 - (-1))i + (1 - 2)j + (1 - 3)k = 2i - j - 2k$
$\overrightarrow{AC} = (2 - (-1))i + (-1 - 2)j + (3 - 3)k = 3i - 3j + 0k$
To find a normal vector to the plane,we calculate the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$:
$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} i & j & k \\ 2 & -1 & -2 \\ 3 & -3 & 0 \end{vmatrix} = i(0 - 6) - j(0 - (-6)) + k(-6 - (-3)) = -6i - 6j - 3k$
Let $\vec{n} = -6i - 6j - 3k$. The magnitude is $|\vec{n}| = \sqrt{(-6)^2 + (-6)^2 + (-3)^2} = \sqrt{36 + 36 + 9} = \sqrt{81} = 9$.
The unit normal vector is $\hat{n} = \pm \frac{\vec{n}}{|\vec{n}|} = \pm \frac{-6i - 6j - 3k}{9} = \pm \left( \frac{-2i - 2j - k}{3} \right) = \pm \left( \frac{2i + 2j + k}{3} \right)$.

(B) The vector equation of a plane at a distance $d$ from the origin and normal to the vector $\vec{n}$ is given by $r \cdot \hat{n} = d$,where $\hat{n}$ is the unit normal vector.
Given $d = 8$ and $\vec{n} = 2\hat{i} + \hat{j} + 2\hat{k}$.
First,calculate the magnitude of $\vec{n}$:
$|\vec{n}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.
Now,find the unit normal vector $\hat{n}$:
$\hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{3}$.
The equation of the plane is $r \cdot \hat{n} = d$:
$r \cdot \left( \frac{2\hat{i} + \hat{j} + 2\hat{k}}{3} \right) = 8$.
Multiplying both sides by $3$,we get:
$r \cdot (2\hat{i} + \hat{j} + 2\hat{k}) = 24$.

(C) Given position vectors of points $P$ and $Q$ are $\vec{p} = 3i + j + 2k$ and $\vec{q} = i - 2j - 4k$.
The vector $\overrightarrow{PQ} = \vec{q} - \vec{p} = (i - 2j - 4k) - (3i + j + 2k) = -2i - 3j - 6k$.
The plane passes through $Q$ and is perpendicular to $\overrightarrow{PQ}$.
The normal vector to the plane is $\vec{n} = \overrightarrow{PQ} = -2i - 3j - 6k$.
The equation of the plane is given by $(r - \vec{q}) \cdot \vec{n} = 0$,which is $r \cdot \vec{n} = \vec{q} \cdot \vec{n}$.
$\vec{q} \cdot \vec{n} = (i - 2j - 4k) \cdot (-2i - 3j - 6k) = (1)(-2) + (-2)(-3) + (-4)(-6) = -2 + 6 + 24 = 28$.
Thus,$r \cdot (-2i - 3j - 6k) = 28$,which can be written as $r \cdot (2i + 3j + 6k) = -28$.

(B) The equation of a plane passing through the line of intersection of the planes $r \cdot a = \lambda$ and $r \cdot b = \mu$ is given by the family of planes equation:
$(r \cdot a - \lambda) + k(r \cdot b - \mu) = 0$
Rearranging the terms,we get:
$r \cdot (a + kb) = \lambda + k\mu$ .....$(i)$
Since this plane passes through the origin,the position vector $r = 0$ must satisfy the equation:
$0 \cdot (a + kb) = \lambda + k\mu$
$0 = \lambda + k\mu$
$k = -\frac{\lambda}{\mu}$
Substituting the value of $k$ into equation $(i)$:
$r \cdot (a - \frac{\lambda}{\mu} b) = \lambda + (-\frac{\lambda}{\mu})\mu$
$r \cdot (\frac{\mu a - \lambda b}{\mu}) = 0$
Multiplying by $\mu$,we get:
$r \cdot (\mu a - \lambda b) = 0$
Which is equivalent to:
$r \cdot (\lambda b - \mu a) = 0$

(B) The equation of a plane parallel to the plane $\vec{r} \cdot (4\hat{i} - 12\hat{j} - 3\hat{k}) - 7 = 0$ is given by $\vec{r} \cdot (4\hat{i} - 12\hat{j} - 3\hat{k}) + \lambda = 0$.
Since this plane passes through the point with position vector $\vec{a} = 2\hat{i} - \hat{j} - 4\hat{k}$,we substitute this into the equation:
$(2\hat{i} - \hat{j} - 4\hat{k}) \cdot (4\hat{i} - 12\hat{j} - 3\hat{k}) + \lambda = 0$
Calculating the dot product:
$(2)(4) + (-1)(-12) + (-4)(-3) + \lambda = 0$
$8 + 12 + 12 + \lambda = 0$
$32 + \lambda = 0$
$\lambda = -32$
Substituting $\lambda = -32$ back into the equation,we get:
$\vec{r} \cdot (4\hat{i} - 12\hat{j} - 3\hat{k}) - 32 = 0$
$\vec{r} \cdot (4\hat{i} - 12\hat{j} - 3\hat{k}) = 32$
Thus,the correct option is $B$.

(A) The required plane passes through a point with position vector $a_1$ and is parallel to the vectors $a_1$ and $a_2$.
If $r$ is the position vector of any point on the plane,then the vectors $(r - a_1)$,$a_1$,and $a_2$ are coplanar.
Therefore,their scalar triple product must be zero: $(r - a_1) \cdot (a_1 \times a_2) = 0$.
Expanding this,we get $r \cdot (a_1 \times a_2) - a_1 \cdot (a_1 \times a_2) = 0$.
Since $[r, a_1, a_2] = r \cdot (a_1 \times a_2)$ and $[a_1, a_1, a_2] = 0$ (as two vectors are identical),we have $[r, a_1, a_2] = 0$.
Hence,the equation of the plane is $[r, a_1, a_2] = 0$.

(C) Given the vector equation of the plane: $r = (1 + \lambda - \mu )i + (2 - \lambda )j + (3 - 2\lambda + 2\mu )k$
This can be rewritten as: $r = (i + 2j + 3k) + \lambda (i - j - 2k) + \mu (-i + 2k)$
This represents a plane passing through the point $a = i + 2j + 3k$ and parallel to the vectors $b = i - j - 2k$ and $c = -i + 2k$.
The normal vector $n$ to the plane is given by the cross product $n = b \times c$:
$n = \begin{vmatrix} i & j & k \\ 1 & -1 & -2 \\ -1 & 0 & 2 \end{vmatrix} = i(-2 - 0) - j(2 - 2) + k(0 - 1) = -2i - k$
The vector equation of the plane is $(r - a) \cdot n = 0$, which implies $r \cdot n = a \cdot n$.
Calculating $a \cdot n$: $(i + 2j + 3k) \cdot (-2i - k) = (1)(-2) + (2)(0) + (3)(-1) = -2 - 3 = -5$.
Thus, $r \cdot (-2i - k) = -5$, or $r \cdot (2i + k) = 5$.
Substituting $r = xi + yj + zk$, we get $(xi + yj + zk) \cdot (2i + k) = 5$, which simplifies to $2x + z = 5$.

(A) The vector equation of the plane passing through three non-collinear points with position vectors $a, b, c$ is given by $r \cdot (a \times b + b \times c + c \times a) = [a, b, c]$.
This equation is in the form $r \cdot n = p$,where $n = a \times b + b \times c + c \times a$ is the normal vector to the plane and $p = [a, b, c]$.
The length of the perpendicular from the origin to the plane $r \cdot n = d$ is given by $\frac{|d|}{|n|}$.
Substituting the values,the length of the perpendicular is $\frac{|[a, b, c]|}{|a \times b + b \times c + c \times a|}$.
Since the scalar triple product $[a, b, c]$ represents the volume of the parallelepiped formed by vectors $a, b, c$,and the denominator represents the magnitude of the normal vector,the correct expression is $\frac{[a, b, c]}{|a \times b + b \times c + c \times a|}$.

(C) The plane passes through the point $a$ and contains the line $r = b + \lambda c$. Therefore,the plane is parallel to the vector $c$ and the vector $(b - a)$.
The normal vector $n$ to the plane is given by the cross product of these two vectors:
$n = (b - a) \times c = b \times c - a \times c = b \times c + c \times a$.
The equation of the plane is $(r - a) \cdot n = 0$,which simplifies to $r \cdot n = a \cdot n$.
Substituting $n = b \times c + c \times a$,we get $r \cdot (b \times c + c \times a) = a \cdot (b \times c + c \times a)$.
Since $a \cdot (b \times c + c \times a) = a \cdot (b \times c) + a \cdot (c \times a) = [a, b, c] + 0 = [a, b, c]$,the equation of the plane is $r \cdot (b \times c + c \times a) = [a, b, c]$.
The length of the perpendicular from the origin $(0, 0, 0)$ to the plane $r \cdot n = d$ is given by $\frac{|d|}{|n|}$.
Here,$d = [a, b, c]$ and $n = b \times c + c \times a$.
Thus,the length of the perpendicular is $\frac{[a, b, c]}{|b \times c + c \times a|}$.

(B) The equations of the two parallel planes are given by $r \cdot n + d_1 = 0$ and $r \cdot n + d_2 = 0$.
Here,$n = i + 2j - 2k$,$d_1 = 5$,and $d_2 = -8$.
The distance $D$ between two parallel planes $r \cdot n + d_1 = 0$ and $r \cdot n + d_2 = 0$ is given by the formula $D = \frac{|d_1 - d_2|}{|n|}$.
First,calculate the magnitude of the normal vector $n$:
$|n| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
Now,substitute the values into the distance formula:
$D = \frac{|5 - (-8)|}{3} = \frac{|5 + 8|}{3} = \frac{13}{3} \text{ units}$.

(A) Let the points be $A(-1, -2, 0)$ and $B(2, 3, 5)$. The vector $\vec{AB} = (2 - (-1))i + (3 - (-2))j + (5 - 0)k = 3i + 5j + 5k$.
The plane is parallel to the line with direction vector $\vec{v} = 2i + 5j - k$.
The normal vector $\vec{n}$ to the plane is given by the cross product of $\vec{AB}$ and $\vec{v}$:
$\vec{n} = \vec{AB} \times \vec{v} = \begin{vmatrix} i & j & k \\ 3 & 5 & 5 \\ 2 & 5 & -1 \end{vmatrix}$
$= i(-5 - 25) - j(-3 - 10) + k(15 - 10) = -30i + 13j + 5k$.
The equation of the plane passing through $(-1, -2, 0)$ is:
$-30(x + 1) + 13(y + 2) + 5(z - 0) = 0$
$-30x - 30 + 13y + 26 + 5z = 0$
$-30x + 13y + 5z = 4$.
In vector form,this is $r \cdot (-30i + 13j + 5k) = 4$.

(A) Let the intercepts of the plane on the $x, y,$ and $z$ axes be $a, b,$ and $c$ respectively.
The coordinates of the vertices are $P(a, 0, 0), Q(0, b, 0),$ and $R(0, 0, c)$.
The centroid of $\Delta PQR$ is given by $(\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$.
Given the centroid is $2i - 5j + 8k$,we have $\frac{a}{3} = 2, \frac{b}{3} = -5, \frac{c}{3} = 8$.
Thus,$a = 6, b = -15, c = 24$.
The intercept form of the equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Substituting the values,we get $\frac{x}{6} + \frac{y}{-15} + \frac{z}{24} = 1$.
Multiplying by the least common multiple of $6, 15,$ and $24$,which is $120$,we get $20x - 8y + 5z = 120$.
In vector form,this is $r \cdot (20i - 8j + 5k) = 120$.

(A) The normal vector $\vec{n}$ to the plane is perpendicular to both vectors $\vec{a}$ and $\vec{b}$.
Thus,$\vec{n} = \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & -1 \\ -3 & 2 & 2 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(6 - 3) + \hat{k}(6 - 0) = 2\hat{i} - 3\hat{j} + 6\hat{k}$.
The equation of the plane passing through point $(x_0, y_0, z_0)$ with normal vector $\vec{n} = (A, B, C)$ is given by $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$.
Substituting the point $A(2, -1, 3)$ and the normal vector $\vec{n} = (2, -3, 6)$:
$2(x - 2) - 3(y + 1) + 6(z - 3) = 0$
$2x - 4 - 3y - 3 + 6z - 18 = 0$
$2x - 3y + 6z - 25 = 0$.

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

(C) The general form of a first-degree equation in three variables $x, y, z$ is given by $Ax + By + Cz + D = 0$,where $A, B, C$ are not all zero.
This equation always represents a plane in three-dimensional space.

(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 direction ratios of the normal as $(A, B, C) = (1, 2, -3)$.
The magnitude of the normal vector is $\sqrt{A^2 + B^2 + C^2} = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
The direction cosines $(l, m, n)$ are given by $\pm \frac{A}{\sqrt{A^2 + B^2 + C^2}}, \pm \frac{B}{\sqrt{A^2 + B^2 + C^2}}, \pm \frac{C}{\sqrt{A^2 + B^2 + C^2}}$.
Taking the negative sign,we get $l = -\frac{1}{\sqrt{14}}$,$m = -\frac{2}{\sqrt{14}}$,and $n = -\frac{-3}{\sqrt{14}} = \frac{3}{\sqrt{14}}$.
Thus,the direction cosines are $(-\frac{1}{\sqrt{14}}, -\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}})$,which matches option $A$.

(B) 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 $\cos \theta = \left| \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$.
Here,the normal vectors are $\vec{n_1} = 3\hat{i} - 4\hat{j} + 5\hat{k}$ and $\vec{n_2} = 2\hat{i} - 1\hat{j} - 2\hat{k}$.
Substituting the values,we get $\cos \theta = \left| \frac{(3)(2) + (-4)(-1) + (5)(-2)}{\sqrt{3^2 + (-4)^2 + 5^2} \sqrt{2^2 + (-1)^2 + (-2)^2}} \right|$.
$\cos \theta = \left| \frac{6 + 4 - 10}{\sqrt{9 + 16 + 25} \sqrt{4 + 1 + 4}} \right| = \left| \frac{0}{\sqrt{50} \sqrt{9}} \right| = 0$.
Since $\cos \theta = 0$,we have $\theta = \cos^{-1}(0) = \frac{\pi}{2}$.

(B) The intercept form of the equation of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Since the plane is parallel to the $y$-axis,it does not intersect the $y$-axis,meaning the intercept on the $y$-axis is infinite,so the term involving $y$ vanishes.
The intercepts on the $x$-axis and $z$-axis are given as $2$ and $3$ respectively.
Thus,the equation of the plane is $\frac{x}{2} + \frac{z}{3} = 1$.
Multiplying the entire equation by $6$,we get $3x + 2z = 6$.

(C) The intercept form of the equation of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the intercepts on the $x, y, z$ axes respectively.
Given intercepts are $a = -6, b = 3, c = 4$.
Substituting these values,we get $\frac{x}{-6} + \frac{y}{3} + \frac{z}{4} = 1$.
To simplify,multiply by the least common multiple of $6, 3, 4$,which is $12$:
$-2x + 4y + 3z = 12$,or $-2x + 4y + 3z - 12 = 0$.
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,$A = -2, B = 4, C = 3, D = -12$.
$d = \frac{|-12|}{\sqrt{(-2)^2 + 4^2 + 3^2}} = \frac{12}{\sqrt{4 + 16 + 9}} = \frac{12}{\sqrt{29}}$.

(C) The general equation of a plane parallel to the $xy$-plane is given by $z = k$,where $k$ is a constant.
Since the plane cuts an intercept of length $3$ from the $z$-axis,the value of $k$ is $3$.
Therefore,the equation of the plane is $z = 3$.

(D) The equations of the planes are $P_1: 3x - 6y + 2z + 5 = 0$ and $P_2: 4x - 12y + 3z - 3 = 0$.
To find the bisector containing the origin,we first make the constant terms positive.
For $P_1$,the constant is $5 > 0$,so we keep it as $3x - 6y + 2z + 5 = 0$.
For $P_2$,the constant is $-3 < 0$,so we multiply by $-1$ to get $-4x + 12y - 3z + 3 = 0$.
The equation of the bisector containing the origin is given by $\frac{3x - 6y + 2z + 5}{\sqrt{3^2 + (-6)^2 + 2^2}} = \frac{-4x + 12y - 3z + 3}{\sqrt{(-4)^2 + 12^2 + (-3)^2}}$.
This simplifies to $\frac{3x - 6y + 2z + 5}{7} = \frac{-4x + 12y - 3z + 3}{13}$.
Cross-multiplying gives $13(3x - 6y + 2z + 5) = 7(-4x + 12y - 3z + 3)$.
$39x - 78y + 26z + 65 = -28x + 84y - 21z + 21$.
Rearranging terms: $67x - 162y + 47z + 44 = 0$.
Since this result is not among the options,the correct choice is $D$.

(A) Two planes $a_1x + b_1y + c_1z = d_1$ and $a_2x + b_2y + c_2z = d_2$ are perpendicular if their normal vectors $\vec{n_1} = (a_1, b_1, c_1)$ and $\vec{n_2} = (a_2, b_2, c_2)$ are perpendicular,i.e.,$\vec{n_1} \cdot \vec{n_2} = 0$.
Here,the normal vectors are $\vec{n_1} = (3, -6, -2)$ and $\vec{n_2} = (2, 1, -k)$.
For the planes to be perpendicular,the dot product of their normal vectors must be zero:
$(3)(2) + (-6)(1) + (-2)(-k) = 0$
$6 - 6 + 2k = 0$
$0 + 2k = 0$
$2k = 0$
$k = 0$.

(B) The equation of a plane passing through the point $(x_1, y_1, z_1)$ is given by $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$.
Substituting the point $(-1, 3, 2)$,we get $A(x + 1) + B(y - 3) + C(z - 2) = 0$ ..... $(i)$.
Since the plane $(i)$ is perpendicular to the planes $x + 2y + 3z = 5$ and $3x + 3y + z = 0$,its normal vector $\vec{n} = (A, B, C)$ must be perpendicular to the normal vectors $\vec{n_1} = (1, 2, 3)$ and $\vec{n_2} = (3, 3, 1)$.
Thus,the normal vector is $\vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & 3 & 1 \end{vmatrix} = \hat{i}(2 - 9) - \hat{j}(1 - 9) + \hat{k}(3 - 6) = -7\hat{i} + 8\hat{j} - 3\hat{k}$.
Comparing this with $(A, B, C)$,we have $A = -7, B = 8, C = -3$.
Substituting these values into equation $(i)$,we get $-7(x + 1) + 8(y - 3) - 3(z - 2) = 0$.
$-7x - 7 + 8y - 24 - 3z + 6 = 0$.
$-7x + 8y - 3z - 25 = 0$,which is equivalent to $7x - 8y + 3z + 25 = 0$.

(A) The given equations of the planes are $x + 2y + 3z + 7 = 0$ and $2x + 4y + 6z + 7 = 0$.
First,we rewrite the second equation to have the same coefficients for $x, y, z$ as the first plane by dividing it by $2$:
$x + 2y + 3z + \frac{7}{2} = 0$.
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 $d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$.
Here,$A = 1, B = 2, C = 3, D_1 = 7$,and $D_2 = \frac{7}{2}$.
Substituting these values into the formula:
$d = \frac{|7 - \frac{7}{2}|}{\sqrt{1^2 + 2^2 + 3^2}} = \frac{|\frac{14 - 7}{2}|}{\sqrt{1 + 4 + 9}} = \frac{\frac{7}{2}}{\sqrt{14}}$.
Simplifying the expression:
$d = \frac{7}{2\sqrt{14}} = \frac{7}{2\sqrt{2} \times \sqrt{7}} = \frac{\sqrt{7}}{2\sqrt{2}}$.
Thus,the correct option is $A$.

(A) The vertices of the triangle $ABC$ are $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.
The vectors representing the sides are $\vec{AB} = (-a, b, 0)$ and $\vec{AC} = (-a, 0, c)$.
The area of the triangle $ABC$ is given by $\Delta = \frac{1}{2} |\vec{AB} \times \vec{AC}|$.
Calculating the cross product: $\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -a & b & 0 \\ -a & 0 & c \end{vmatrix} = \hat{i}(bc) - \hat{j}(-ac) + \hat{k}(ab) = (bc, ac, ab)$.
The magnitude is $|\vec{AB} \times \vec{AC}| = \sqrt{(bc)^2 + (ac)^2 + (ab)^2} = \sqrt{b^2c^2 + a^2c^2 + a^2b^2}$.
Therefore,the area is $\Delta = \frac{1}{2} \sqrt{b^2c^2 + c^2a^2 + a^2b^2}$.

(C) The equation of a plane parallel to $x - 2y + 2z = 5$ is of the form $x - 2y + 2z + k = 0$ ... $(i)$.
The distance $d$ 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 point $(1, 2, 3)$ and distance $d = 1$,we have:
$\frac{|1 - 2(2) + 2(3) + k|}{\sqrt{1^2 + (-2)^2 + 2^2}} = 1$
$\frac{|1 - 4 + 6 + k|}{\sqrt{1 + 4 + 4}} = 1$
$\frac{|k + 3|}{\sqrt{9}} = 1$
$|k + 3| = 3$
This implies $k + 3 = 3$ or $k + 3 = -3$.
Case $1$: $k = 0$. Substituting into $(i)$,we get $x - 2y + 2z = 0$.
Case $2$: $k = -6$. Substituting into $(i)$,we get $x - 2y + 2z - 6 = 0$,which is $x - 2y + 2z = 6$.
Comparing with the given options,$x - 2y + 2z = 6$ is the correct equation.

(D) The equation of a plane parallel to $ax + by + cz + d = 0$ is given by $ax + by + cz + k = 0$.
Therefore,the plane parallel to $2x + 3y - 4z = 0$ is $2x + 3y - 4z + k = 0$.....$(i)$
Since the plane $(i)$ passes through the point $(1, 2, 3)$,we substitute these coordinates into the equation:
$2(1) + 3(2) - 4(3) + k = 0$
$2 + 6 - 12 + k = 0$
$8 - 12 + k = 0$
$-4 + k = 0$
$k = 4$
Substituting the value of $k$ back into equation $(i)$,we get the required plane equation:
$2x + 3y - 4z + 4 = 0$.

(B) Let the moving point be $P(x, y, z)$.
Given that 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 $\vec{n} = 2\hat{i} + 2\hat{j} + 2\hat{k}$.
Since the components of the normal vector are equal,the normal is equally inclined to the coordinate axes.

(A) In a three-dimensional Cartesian coordinate system,any point in the $yz$-plane has its $x$-coordinate equal to $0$.
Therefore,the equation representing the $yz$-plane is $x = 0$.

(D) The equation of the first plane is $2x - y + z = 6$,so its normal vector is $\vec{n_1} = 2\hat{i} - \hat{j} + \hat{k}$.
The equation of the second plane is $x + y + 2z = 7$,so its normal vector is $\vec{n_2} = \hat{i} + \hat{j} + 2\hat{k}$.
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: $\vec{n_1} \cdot \vec{n_2} = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3$.
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,$\theta = \cos^{-1}(\frac{1}{2}) = 60^o$.

(A) Let the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane be $P(2, 4, -3)$.
Since $OP$ is the normal to the plane,the direction ratios of the normal are $(2 - 0, 4 - 0, -3 - 0) = (2, 4, -3)$.
The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal vector $(a, b, c)$ is given by $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$.
Substituting the point $P(2, 4, -3)$ and the normal vector $(2, 4, -3)$,we get:
$2(x - 2) + 4(y - 4) - 3(z - (-3)) = 0$
$2(x - 2) + 4(y - 4) - 3(z + 3) = 0$
$2x - 4 + 4y - 16 - 3z - 9 = 0$
$2x + 4y - 3z - 29 = 0$
$2x + 4y - 3z = 29$.

(A) The normal vector $\vec{n}$ to the plane is the vector joining the points $(3, 4, -1)$ and $(2, -1, 5)$.
$\vec{n} = (2 - 3)\hat{i} + (-1 - 4)\hat{j} + (5 - (-1))\hat{k} = -1\hat{i} - 5\hat{j} + 6\hat{k}$.
Since the plane is normal to this line,the normal vector to the plane is $\vec{n} = -\hat{i} - 5\hat{j} + 6\hat{k}$ or $\vec{n} = \hat{i} + 5\hat{j} - 6\hat{k}$.
The equation of a plane passing through $(x_0, y_0, z_0)$ with normal vector $a\hat{i} + b\hat{j} + c\hat{k}$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.
Substituting $(x_0, y_0, z_0) = (2, -3, 1)$ and $(a, b, c) = (1, 5, -6)$:
$1(x - 2) + 5(y - (-3)) - 6(z - 1) = 0$
$x - 2 + 5y + 15 - 6z + 6 = 0$
$x + 5y - 6z + 19 = 0$.

(A) Let the equation of the plane be $a(x - 1) + b(y - 1) + c(z - 1) = 0$.
Since the plane passes through $(1, -1, 1)$,we have $a(1 - 1) + b(-1 - 1) + c(1 - 1) = 0$,which implies $-2b = 0$,so $b = 0$.
Now the equation becomes $a(x - 1) + c(z - 1) = 0$.
Since it also passes through $(-7, -3, -5)$,we have $a(-7 - 1) + c(-5 - 1) = 0$,which gives $-8a - 6c = 0$,or $8a = -6c$,which simplifies to $a = -\frac{3}{4}c$.
Substituting $a = -\frac{3}{4}c$ into the equation,we get $-\frac{3}{4}c(x - 1) + c(z - 1) = 0$.
Dividing by $c$ (assuming $c \neq 0$) and multiplying by $-4$,we get $3(x - 1) - 4(z - 1) = 0$,which simplifies to $3x - 3 - 4z + 4 = 0$,or $3x - 4z + 1 = 0$.

(A) Let $Q$ be the image of the point $P(1, 3, 4)$ in the given plane $2x - y + z + 3 = 0$. The line $PQ$ is normal to the plane.
The direction ratios of the normal to the plane are $2, -1, 1$.
Since $PQ$ passes through $P(1, 3, 4)$ and is parallel to the normal vector,the equation of the line $PQ$ is $\frac{x - 1}{2} = \frac{y - 3}{-1} = \frac{z - 4}{1} = r$.
Any point on this line is given by $(2r + 1, -r + 3, r + 4)$. Let this be $Q$.
The midpoint $R$ of $PQ$ is $\left( \frac{2r + 1 + 1}{2}, \frac{-r + 3 + 3}{2}, \frac{r + 4 + 4}{2} \right) = \left( r + 1, \frac{-r + 6}{2}, \frac{r + 8}{2} \right)$.
Since $R$ lies on the plane $2x - y + z + 3 = 0$,we substitute the coordinates of $R$ into the plane equation:
$2(r + 1) - \left( \frac{-r + 6}{2} \right) + \left( \frac{r + 8}{2} \right) + 3 = 0$
Multiplying by $2$:
$4(r + 1) - (-r + 6) + (r + 8) + 6 = 0$
$4r + 4 + r - 6 + r + 8 + 6 = 0$
$6r + 12 = 0 \implies r = -2$.
Substituting $r = -2$ into the coordinates of $Q$:
$x = 2(-2) + 1 = -3$
$y = -(-2) + 3 = 5$
$z = -2 + 4 = 2$
Thus,the image point is $(-3, 5, 2)$.

(C) The equations of the given planes are $2x - 2y + z + 3 = 0$ and $4x - 4y + 2z + 5 = 0$.
First,we normalize the second equation so that the coefficients of $x, y, z$ are the same as in the first equation. Dividing the second equation by $2$,we get $2x - 2y + z + \frac{5}{2} = 0$.
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 $d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$.
Here,$A = 2, B = -2, C = 1, D_1 = 3$,and $D_2 = \frac{5}{2}$.
Substituting these values,we get $d = \frac{|3 - \frac{5}{2}|}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{|\frac{6-5}{2}|}{\sqrt{4 + 4 + 1}} = \frac{\frac{1}{2}}{\sqrt{9}} = \frac{\frac{1}{2}}{3} = \frac{1}{6}$.

(D) The normal vectors to the planes $ax + by + cz + d = 0$ and $a'x + b'y + c'z + d' = 0$ are $\vec{n_1} = a\hat{i} + b\hat{j} + c\hat{k}$ and $\vec{n_2} = a'\hat{i} + b'\hat{j} + c'\hat{k}$ respectively.
Two planes are mutually perpendicular if and only if their normal vectors are perpendicular.
Two vectors are perpendicular if their dot product is zero,i.e.,$\vec{n_1} \cdot \vec{n_2} = 0$.
Calculating the dot product: $(a\hat{i} + b\hat{j} + c\hat{k}) \cdot (a'\hat{i} + b'\hat{j} + c'\hat{k}) = aa' + bb' + cc' = 0$.
Therefore,the correct condition is $aa' + bb' + cc' = 0$.

(B) The angle between two planes is defined as the angle between their normal vectors. If the equations of the two planes are $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$,their normals are $\vec{n_1} = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$ and $\vec{n_2} = a_2\hat{i} + b_2\hat{j} + c_2\hat{k}$. The angle $\theta$ between the planes is given by $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.

(A) The given equation is $3y + 4z = 0$.
This is a linear equation in $x, y, z$ of the form $Ax + By + Cz + D = 0$,where $A=0, B=3, C=4, D=0$.
Since the equation is of the form $By + Cz = 0$,it represents a plane.
For a plane to contain the $x$-axis,the coordinates of any point on the $x$-axis (which are of the form $(k, 0, 0)$) must satisfy the equation.
Substituting $(k, 0, 0)$ into $3y + 4z = 0$,we get $3(0) + 4(0) = 0$,which is $0 = 0$.
Since the equation is satisfied for all values of $k$,the plane contains the $x$-axis.

(A) Let the coordinates of the points where the plane meets the axes be $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.
The centroid of triangle $ABC$ is given by $(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}) = (\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$.
Given that the centroid is $(\alpha, \beta, \gamma)$,we have $\frac{a}{3} = \alpha$,$\frac{b}{3} = \beta$,and $\frac{c}{3} = \gamma$.
Thus,$a = 3\alpha$,$b = 3\beta$,and $c = 3\gamma$.
The intercept form of the equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Substituting the values of $a, b, c$,we get $\frac{x}{3\alpha} + \frac{y}{3\beta} + \frac{z}{3\gamma} = 1$.
Multiplying both sides by $3$,we obtain $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3$.

(A) Two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are perpendicular if and only if $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
Given the equations of the planes are $3x - 2y + 2z + 17 = 0$ and $4x + 3y - kz - 25 = 0$.
Here,$a_1 = 3, b_1 = -2, c_1 = 2$ and $a_2 = 4, b_2 = 3, c_2 = -k$.
Substituting these values into the condition for perpendicularity:
$(3)(4) + (-2)(3) + (2)(-k) = 0$
$12 - 6 - 2k = 0$
$6 - 2k = 0$
$2k = 6$
$k = 3$.

(C) The normal vector to the plane is the vector $\vec{OA}$.
Since $O = (0, 0, 0)$ and $A = (a, b, c)$,the direction ratios of the normal vector $\vec{OA}$ are $(a - 0, b - 0, c - 0)$,which are $a, b, c$.
The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal vector $(A, B, C)$ is given by $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$.
Substituting the point $A(a, b, c)$ and the normal vector $(a, b, c)$ into the formula:
$a(x - a) + b(y - b) + c(z - c) = 0$.
Thus,the correct option is $C$.

(A) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Since the plane is at a constant distance $p$ from the origin $(0, 0, 0)$,we have $p = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$.
Squaring both sides,we get $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{p^2}$ ..... $(i)$.
The coordinates of $A, B, C$ are $(a, 0, 0), (0, b, 0)$ and $(0, 0, c)$ respectively.
The centroid $(x, y, z)$ of the tetrahedron $OABC$ is given by $x = \frac{0+a+0+0}{4} = \frac{a}{4}$,$y = \frac{0+0+b+0}{4} = \frac{b}{4}$,and $z = \frac{0+0+0+c}{4} = \frac{c}{4}$.
Thus,$a = 4x, b = 4y, c = 4z$.
Substituting these into equation $(i)$,we get $\frac{1}{(4x)^2} + \frac{1}{(4y)^2} + \frac{1}{(4z)^2} = \frac{1}{p^2}$.
$\frac{1}{16x^2} + \frac{1}{16y^2} + \frac{1}{16z^2} = \frac{1}{p^2}$.
Multiplying by $16$,we get $x^{-2} + y^{-2} + z^{-2} = 16p^{-2}$.

(D) The equation of the plane is $ax + by + cz = 1$.
To find the intersection with the $x$-axis,set $y = 0$ and $z = 0$,which gives $ax = 1$,so $x = \frac{1}{a}$. Thus,point $A = (\frac{1}{a}, 0, 0)$.
To find the intersection with the $y$-axis,set $x = 0$ and $z = 0$,which gives $by = 1$,so $y = \frac{1}{b}$. Thus,point $B = (0, \frac{1}{b}, 0)$.
To find the intersection with the $z$-axis,set $x = 0$ and $y = 0$,which gives $cz = 1$,so $z = \frac{1}{c}$. Thus,point $C = (0, 0, \frac{1}{c})$.
The centroid of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$.
Substituting the coordinates of $A, B$,and $C$:
Centroid $= (\frac{\frac{1}{a} + 0 + 0}{3}, \frac{0 + \frac{1}{b} + 0}{3}, \frac{0 + 0 + \frac{1}{c}}{3}) = (\frac{1}{3a}, \frac{1}{3b}, \frac{1}{3c})$.
Therefore,the correct option is $D$.

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

(D) The general equation of a plane parallel to $Ax + By + Cz = D$ is given by $Ax + By + Cz = k$.
Since the plane is parallel to $x + 2y + 4z = 5$,its equation must be of the form $x + 2y + 4z = k$.
This plane passes through the point $(2, 3, 4)$,so we substitute these coordinates into the equation to find $k$:
$2 + 2(3) + 4(4) = k$
$2 + 6 + 16 = k$
$k = 24$
Therefore,the equation of the plane is $x + 2y + 4z = 24$.

(D) The equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 3$.
To find the intersection with the $x$-axis,set $y=0$ and $z=0$: $\frac{x}{a} = 3 \implies x = 3a$. So,$A = (3a, 0, 0)$.
To find the intersection with the $y$-axis,set $x=0$ and $z=0$: $\frac{y}{b} = 3 \implies y = 3b$. So,$B = (0, 3b, 0)$.
To find the intersection with the $z$-axis,set $x=0$ and $y=0$: $\frac{z}{c} = 3 \implies z = 3c$. So,$C = (0, 0, 3c)$.
The centroid of a triangle with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2),$ and $(x_3, y_3, z_3)$ is given by $\left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3} \right)$.
Substituting the coordinates of $A, B,$ and $C$:
Centroid $= \left( \frac{3a+0+0}{3}, \frac{0+3b+0}{3}, \frac{0+0+3c}{3} \right) = (a, b, c)$.

(B) The equation of a plane,when the direction ratios $(a, b, c)$ of the normal and the length of the perpendicular $p$ from the origin are given,is $ax + by + cz = p \sqrt{a^2 + b^2 + c^2}$.
Given,$(a, b, c) = (-3, 2, 6)$ and $p = 7$.
Substituting these values into the formula:
$-3x + 2y + 6z = 7 \sqrt{(-3)^2 + 2^2 + 6^2}$
$-3x + 2y + 6z = 7 \sqrt{9 + 4 + 36}$
$-3x + 2y + 6z = 7 \sqrt{49}$
$-3x + 2y + 6z = 7 \times 7$
$-3x + 2y + 6z = 49$
Thus,the equation is $-3x + 2y + 6z - 49 = 0$.

(A) Given the plane equation is $x - 3y + 5z = d$.
Since the plane passes through the point $(1, 2, 4)$,we substitute these coordinates into the equation:
$1 - 3(2) + 5(4) = d$
$1 - 6 + 20 = d$
$d = 15$
Thus,the equation of the plane is $x - 3y + 5z = 15$.
To find the intercepts,we divide the entire equation by $15$:
$\frac{x}{15} + \frac{-3y}{15} + \frac{5z}{15} = \frac{15}{15}$
$\frac{x}{15} + \frac{y}{-5} + \frac{z}{3} = 1$
Comparing this with the intercept form of a plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,the intercepts on the $x, y, z$ axes are $15, -5, 3$ respectively.

(D) Two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are perpendicular if $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
Here,the normal vectors to the planes are $\vec{n_1} = (1, 2, k)$ and $\vec{n_2} = (2, 1, -2)$.
Since the planes are at right angles,their normal vectors must be perpendicular,so their dot product is zero:
$(1)(2) + (2)(1) + (k)(-2) = 0$.
$2 + 2 - 2k = 0$.
$4 - 2k = 0$.
$2k = 4$.
$k = 2$.

(B) When two planes intersect,they share a common line of intersection.
Since the planes meet at this line,every point on the line of intersection lies on both planes.
Therefore,the shortest distance between two intersecting planes is $0$.
Since $\cos {90^o} = 0$,the correct option is $(b)$.