Line and Plane Questions in English

(A) The line is given by $r = (i + 2j - k) + \lambda (i - j + k)$. The direction vector of the line is $b = i - j + k$.
The plane is given by $r \cdot (2i - j + k) = 4$. The normal vector to the plane is $n = 2i - j + k$.
The angle $\theta$ between the line and the normal to the plane is given by the formula $\cos \theta = \frac{|b \cdot n|}{|b| |n|}$.
Calculating the dot product: $b \cdot n = (1)(2) + (-1)(-1) + (1)(1) = 2 + 1 + 1 = 4$.
Calculating the magnitudes: $|b| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$ and $|n| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$.
Thus,$\cos \theta = \frac{4}{\sqrt{3} \sqrt{6}} = \frac{4}{\sqrt{18}} = \frac{4}{3\sqrt{2}} = \frac{2\sqrt{2}}{3}$.
Therefore,$\theta = \cos^{-1} \left( \frac{2\sqrt{2}}{3} \right)$.
However,the question asks for the angle between the line and the normal. If we denote the angle between the line and the plane as $\alpha$,then $\sin \alpha = \cos \theta = \frac{2\sqrt{2}}{3}$.
Given the options provided,the intended answer is $\sin^{-1} \left( \frac{2\sqrt{2}}{3} \right)$.

(A) The perpendicular distance of a point $P$ with position vector $\vec{a}$ from the plane $\vec{r} \cdot \vec{n} = d$ is given by the formula: $D = \frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|}$.
Given the point $\vec{a} = 2i + j - k$,the plane equation $\vec{r} \cdot (i - 2j + 4k) = 9$,we identify $\vec{n} = i - 2j + 4k$ and $d = 9$.
First,calculate the dot product $\vec{a} \cdot \vec{n} = (2)(1) + (1)(-2) + (-1)(4) = 2 - 2 - 4 = -4$.
Next,calculate the magnitude of the normal vector $|\vec{n}| = \sqrt{1^2 + (-2)^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21}$.
Substitute these values into the distance formula: $D = \frac{|-4 - 9|}{\sqrt{21}} = \frac{|-13|}{\sqrt{21}} = \frac{13}{\sqrt{21}}$.

(B) The given equation of the plane is $r \cdot (i + 2j + 2k) = 15$. The normal vector to the plane is $n = i + 2j + 2k$.
The sphere is given by $|r - (j + 2k)| = 4$,which has its centre at $C_0 = (0, 1, 2)$ and radius $R = 4$.
The centre of the circle is the projection of the centre of the sphere $C_0$ onto the plane.
The line passing through $C_0$ and perpendicular to the plane is given by $r = (j + 2k) + \lambda (i + 2j + 2k) = \lambda i + (1 + 2\lambda) j + (2 + 2\lambda) k$.
Substituting this into the plane equation: $(\lambda i + (1 + 2\lambda) j + (2 + 2\lambda) k) \cdot (i + 2j + 2k) = 15$.
$\lambda + 2(1 + 2\lambda) + 2(2 + 2\lambda) = 15$.
$\lambda + 2 + 4\lambda + 4 + 4\lambda = 15 \Rightarrow 9\lambda + 6 = 15 \Rightarrow 9\lambda = 9 \Rightarrow \lambda = 1$.
Substituting $\lambda = 1$ back into the line equation,the centre of the circle is $1i + (1 + 2(1))j + (2 + 2(1))k = i + 3j + 4k$,which corresponds to the coordinates $(1, 3, 4)$.

(C) The position vectors of the given points are $a = i - j + 3k$ and $b = 3i + 3j + 3k$.
The equation of the plane is $r \cdot (5i + 2j - 7k) + 9 = 0$. Let $f(r) = r \cdot (5i + 2j - 7k) + 9$.
For point $A$,we calculate $f(a) = (i - j + 3k) \cdot (5i + 2j - 7k) + 9 = (1)(5) + (-1)(2) + (3)(-7) + 9 = 5 - 2 - 21 + 9 = -9$.
Since $f(a) < 0$,point $A$ lies on one side of the plane.
For point $B$,we calculate $f(b) = (3i + 3j + 3k) \cdot (5i + 2j - 7k) + 9 = (3)(5) + (3)(2) + (3)(-7) + 9 = 15 + 6 - 21 + 9 = 9$.
Since $f(b) > 0$,point $B$ lies on the other side of the plane.
Because $f(a)$ and $f(b)$ have opposite signs,the points $A$ and $B$ lie on the opposite sides of the plane.

(A) The equation of a plane passing through the line of intersection of the planes $P_1: r \cdot n_1 = d_1$ and $P_2: r \cdot n_2 = d_2$ is given by $(r \cdot n_1 - d_1) + \lambda (r \cdot n_2 - d_2) = 0$.
Here,$n_1 = i + 3j - k$,$d_1 = 0$,$n_2 = j + 2k$,and $d_2 = 0$.
So,the equation is $(r \cdot (i + 3j - k)) + \lambda (r \cdot (j + 2k)) = 0$ ..... $(i)$
This plane passes through the point $(2, 1, -1)$,which corresponds to the position vector $a = 2i + j - k$.
Substituting $r = 2i + j - k$ into equation $(i)$:
$((2i + j - k) \cdot (i + 3j - k)) + \lambda ((2i + j - k) \cdot (j + 2k)) = 0$
$(2(1) + 1(3) + (-1)(-1)) + \lambda (2(0) + 1(1) + (-1)(2)) = 0$
$(2 + 3 + 1) + \lambda (0 + 1 - 2) = 0$
$6 + \lambda (-1) = 0 \Rightarrow \lambda = 6$
Substituting $\lambda = 6$ back into equation $(i)$:
$r \cdot (i + 3j - k) + 6(r \cdot (j + 2k)) = 0$
$r \cdot (i + 3j - k + 6j + 12k) = 0$
$r \cdot (i + 9j + 11k) = 0$
Thus,the correct option is $A$.

(B) The line of intersection of the planes $r \cdot (3i - j + k) = 1$ and $r \cdot (i + 4j - 2k) = 2$ is parallel to the vector $n = n_1 \times n_2$,where $n_1 = 3i - j + k$ and $n_2 = i + 4j - 2k$.
Calculating the cross product: $n = n_1 \times n_2 = \begin{vmatrix} i & j & k \\ 3 & -1 & 1 \\ 1 & 4 & -2 \end{vmatrix} = i(2 - 4) - j(-6 - 1) + k(12 + 1) = -2i + 7j + 13k$.
The plane is perpendicular to this line,so the normal vector to the plane is $n = -2i + 7j + 13k$.
The plane passes through the point $a = i + 2j - k$.
The equation of the plane is $(r - a) \cdot n = 0$,which is $r \cdot n = a \cdot n$.
$r \cdot (-2i + 7j + 13k) = (i + 2j - k) \cdot (-2i + 7j + 13k) = -2 + 14 - 13 = -1$.
Multiplying by $-1$,we get $r \cdot (2i - 7j - 13k) = 1$.

(B) The given lines are $r = (i + j) + \lambda (i + 2j - k)$ and $r = (i + j) + \mu (-i + j - 2k)$.
Both lines pass through the point $a = i + j$.
The lines are parallel to the vectors $b = i + 2j - k$ and $c = -i + j - 2k$ respectively.
The normal vector $n$ to the plane is given by the cross product $n = b \times c$.
$n = \begin{vmatrix} i & j & k \\ 1 & 2 & -1 \\ -1 & 1 & -2 \end{vmatrix} = i(-4 + 1) - j(-2 - 1) + k(1 + 2) = -3i + 3j + 3k$.
We can simplify the normal vector by dividing by $-3$,so $n' = i - j - k$.
The equation of the plane is $(r - a) \cdot n' = 0$.
Substituting the values,$(r - (i + j)) \cdot (i - j - k) = 0$.
$r \cdot (i - j - k) - (i + j) \cdot (i - j - k) = 0$.
$r \cdot (i - j - k) - (1 - 1 + 0) = 0$.
$r \cdot (i - j - k) = 0$.

(C) The equation of the line is $r = a + \lambda b$,where $b = -i + j + k$.
The equation of the plane is $r \cdot n = d$,where $n = 3i + 2j - k$.
The angle $\theta$ between a line and a plane is given by $\sin \theta = \frac{|b \cdot n|}{|b||n|}$.
Here,$b \cdot n = (-1)(3) + (1)(2) + (1)(-1) = -3 + 2 - 1 = -2$.
$|b| = \sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{3}$.
$|n| = \sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$.
Thus,$\sin \theta = \frac{|-2|}{\sqrt{3} \cdot \sqrt{14}} = \frac{2}{\sqrt{42}}$.
Therefore,$\theta = \sin^{-1}\left(\frac{2}{\sqrt{42}}\right)$.
Note: Since the options provided include $\sin^{-1}\left(\frac{-2}{\sqrt{42}}\right)$,we select the option that matches the magnitude calculation.

(B) The vector equation of the line passing through points $A = i - 2j + k$ and $B = -2j + 3k$ is given by $r = A + \lambda(B - A)$.
$r = (i - 2j + k) + \lambda(-i + 2k) = (1 - \lambda)i - 2j + (1 + 2\lambda)k$ ... $(i)$
The plane passes through the origin $(0, 0, 0)$,$P = 4j$,and $Q = 2i + k$. The normal vector $n$ to the plane is given by $n = P \times Q = (4j) \times (2i + k) = 8(j \times i) + 4(j \times k) = -8k + 4i = 4i - 8k$.
The equation of the plane is $r \cdot (4i - 8k) = 0$ ... $(ii)$
Substituting the point from $(i)$ into $(ii)$:
$((1 - \lambda)i - 2j + (1 + 2\lambda)k) \cdot (4i - 8k) = 0$
$4(1 - \lambda) - 8(1 + 2\lambda) = 0$
$4 - 4\lambda - 8 - 16\lambda = 0$
$-4 - 20\lambda = 0 \Rightarrow \lambda = -\frac{1}{5}$
Substituting $\lambda = -\frac{1}{5}$ into $(i)$:
$r = (i - 2j + k) - \frac{1}{5}(-i + 2k) = i - 2j + k + \frac{1}{5}i - \frac{2}{5}k = \frac{6}{5}i - 2j + \frac{3}{5}k = \frac{1}{5}(6i - 10j + 3k)$.

(A) The given line is $r = (i + j) + \lambda (2i + j + 4k)$.
This line passes through the point $P(1, 1, 0)$ and is parallel to the vector $v = 2i + j + 4k$.
$A$ plane containing this line must satisfy the condition that the point $P(1, 1, 0)$ lies on the plane and the normal vector $n$ of the plane is perpendicular to the direction vector $v$ of the line.
Let the equation of the plane be $r \cdot n = d$.
Checking option $A$: $r \cdot (i + 2j - k) = 3$.
$1$. Check if point $P(1, 1, 0)$ lies on the plane: $(i + j) \cdot (i + 2j - k) = (1)(1) + (1)(2) + (0)(-1) = 1 + 2 = 3$. This is satisfied.
$2$. Check if the line is parallel to the plane: The normal vector $n = i + 2j - k$ must be perpendicular to the line's direction vector $v = 2i + j + 4k$.
$n \cdot v = (i + 2j - k) \cdot (2i + j + 4k) = (1)(2) + (2)(1) + (-1)(4) = 2 + 2 - 4 = 0$.
Since the dot product is $0$,the line is parallel to the plane.
Thus,the plane $r \cdot (i + 2j - k) = 3$ contains the given line.

(D) The line is given by $r = a + \lambda b$,where $a = 2i - 2j + 3k$ and $b = i - j + 4k$.
The plane is given by $r \cdot n = d$,where $n = i + 5j + k$ and $d = 5$.
First,check if the line is parallel to the plane by calculating $b \cdot n$:
$b \cdot n = (i - j + 4k) \cdot (i + 5j + k) = (1)(1) + (-1)(5) + (4)(1) = 1 - 5 + 4 = 0$.
Since $b \cdot n = 0$,the line is parallel to the plane.
The distance between a parallel line and a plane is given by the formula:
$Distance = \left| \frac{d - a \cdot n}{|n|} \right|$
$a \cdot n = (2i - 2j + 3k) \cdot (i + 5j + k) = (2)(1) + (-2)(5) + (3)(1) = 2 - 10 + 3 = -5$.
$|n| = \sqrt{1^2 + 5^2 + 1^2} = \sqrt{1 + 25 + 1} = \sqrt{27} = 3\sqrt{3}$.
$Distance = \left| \frac{5 - (-5)}{3\sqrt{3}} \right| = \left| \frac{10}{3\sqrt{3}} \right| = \frac{10}{3\sqrt{3}}$.

(C) Let $Q$ be the image of the point $P(\vec{i} + 3\vec{k})$ in the plane $\vec{r} \cdot (\vec{i} + \vec{j} + \vec{k}) = 1$. Then the line $PQ$ is normal to the plane.
Since $PQ$ passes through $P$ and is normal to the given plane,the equation of line $PQ$ is $\vec{r} = (\vec{i} + 3\vec{k}) + \lambda(\vec{i} + \vec{j} + \vec{k})$.
Since $Q$ lies on the line $PQ$,let the position vector of $Q$ be $(1 + \lambda)\vec{i} + \lambda\vec{j} + (3 + \lambda)\vec{k}$.
Let $R$ be the midpoint of $PQ$. The position vector of $R$ is $\frac{(\vec{i} + 3\vec{k}) + ((1 + \lambda)\vec{i} + \lambda\vec{j} + (3 + \lambda)\vec{k})}{2} = (\frac{\lambda + 2}{2})\vec{i} + (\frac{\lambda}{2})\vec{j} + (\frac{\lambda + 6}{2})\vec{k}$.
Since $R$ lies on the plane $\vec{r} \cdot (\vec{i} + \vec{j} + \vec{k}) = 1$,we have:
$(\frac{\lambda + 2}{2} + \frac{\lambda}{2} + \frac{\lambda + 6}{2}) = 1$
$\frac{3\lambda + 8}{2} = 1$
$3\lambda + 8 = 2$
$3\lambda = -6 \implies \lambda = -2$.
Substituting $\lambda = -2$ into the expression for $Q$:
$Q = (1 - 2)\vec{i} + (-2)\vec{j} + (3 - 2)\vec{k} = -\vec{i} - 2\vec{j} + \vec{k}$.

(B) The equation of the line is $r = (i - j + k) + t(i + j - k)$.
Any point on this line can be represented as $(1 + t, -1 + t, 1 - t)$.
The equation of the plane is $r \cdot (i + j + k) = 5$,which in Cartesian form is $x + y + z = 5$.
Since the point lies on the plane,it must satisfy the plane equation:
$(1 + t) + (-1 + t) + (1 - t) = 5$
$1 + t = 5$
$t = 4$
Substituting $t = 4$ into the point coordinates:
$x = 1 + 4 = 5$
$y = -1 + 4 = 3$
$z = 1 - 4 = -3$
Thus,the point is $(5, 3, -3)$,and its position vector is $5i + 3j - 3k$.

(B) The line of intersection of the planes $r \cdot (i - 3j + k) = 1$ and $r \cdot (2i + 5j - 3k) = 2$ is perpendicular to each of the normal vectors $n_1 = i - 3j + k$ and $n_2 = 2i + 5j - 3k$.
Therefore,the line is parallel to the vector $n_1 \times n_2$.
$n_1 \times n_2 = \begin{vmatrix} i & j & k \\ 1 & -3 & 1 \\ 2 & 5 & -3 \end{vmatrix}$
$= i((-3)(-3) - (1)(5)) - j((1)(-3) - (1)(2)) + k((1)(5) - (-3)(2))$
$= i(9 - 5) - j(-3 - 2) + k(5 + 6)$
$= 4i + 5j + 11k$.

(D) The equation of the $XOZ$ plane is $y = 0$.
Let the ratio in which the plane $y = 0$ divides the line segment joining the points $A(1, -1, 5)$ and $B(2, 3, 4)$ be $\lambda : 1$.
Using the section formula,the $y$-coordinate of the point of division is given by:
$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$
Substituting the values,we get:
$0 = \frac{\lambda(3) + 1(-1)}{\lambda + 1}$
$0 = 3\lambda - 1$
$3\lambda = 1$
$\lambda = \frac{1}{3}$
Thus,the ratio is $\frac{1}{3} : 1$.

(A) The line is the intersection of two planes: $x - y + z - 5 = 0$ and $x - 3y - 6 = 0$.
Let the direction ratios of the line be $(l, m, n)$.
The normal vectors to the planes are $\vec{n_1} = (1, -1, 1)$ and $\vec{n_2} = (1, -3, 0)$.
The direction of the line is perpendicular to both normals,so it is given by the cross product $\vec{v} = \vec{n_1} \times \vec{n_2}$.
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & -3 & 0 \end{vmatrix} = \hat{i}(0 - (-3)) - \hat{j}(0 - 1) + \hat{k}(-3 - (-1))$
$\vec{v} = 3\hat{i} + 1\hat{j} - 2\hat{k}$.
Thus,the direction ratios are $(3, 1, -2)$.
Therefore,the correct option is $A$.

(C) Two lines $\frac{x - x_1}{l_1} = \frac{y - y_1}{m_1} = \frac{z - z_1}{n_1}$ and $\frac{x - x_2}{l_2} = \frac{y - y_2}{m_2} = \frac{z - z_2}{n_2}$ are coplanar if $\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$.
Here,$(x_1, y_1, z_1) = (2, 3, 4)$ and $(x_2, y_2, z_2) = (1, 4, 5)$.
Thus,$x_2 - x_1 = -1$,$y_2 - y_1 = 1$,$z_2 - z_1 = 1$.
The condition becomes $\begin{vmatrix} -1 & 1 & 1 \\ 1 & 1 & -k \\ k & 2 & 1 \end{vmatrix} = 0$.
Expanding the determinant:
$-1(1 + 2k) - 1(1 + k^2) + 1(2 - k) = 0$
$-1 - 2k - 1 - k^2 + 2 - k = 0$
$-k^2 - 3k = 0$
$k(k + 3) = 0$
Therefore,$k = 0$ or $k = -3$.

(C) The distance of the point $P(1, 1, 1)$ from the origin $O(0, 0, 0)$ is given by $d_1 = \sqrt{(1-0)^2 + (1-0)^2 + (1-0)^2} = \sqrt{3}$.
The distance of the point $P(1, 1, 1)$ from the plane $x - y + z + k = 0$ is given by $d_2 = \frac{|1 - 1 + 1 + k|}{\sqrt{1^2 + (-1)^2 + 1^2}} = \frac{|1 + k|}{\sqrt{3}}$.
The product of the distances is $d_1 \times d_2 = 5$.
Substituting the values,we get $\sqrt{3} \times \frac{|1 + k|}{\sqrt{3}} = 5$.
This simplifies to $|1 + k| = 5$.
Case $1$: $1 + k = 5 \implies k = 4$.
Case $2$: $1 + k = -5 \implies k = -6$.
Comparing with the given options,the correct value is $k = 4$.

(A) The distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by the formula:
$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$
Given point $(x_1, y_1, z_1) = (2, 3, 4)$ and plane $3x - 6y + 2z + 11 = 0$.
Substituting the values:
$d = \frac{|3(2) - 6(3) + 2(4) + 11|}{\sqrt{3^2 + (-6)^2 + 2^2}}$
$d = \frac{|6 - 18 + 8 + 11|}{\sqrt{9 + 36 + 4}}$
$d = \frac{|7|}{\sqrt{49}}$
$d = \frac{7}{7} = 1$
Thus,the distance is $1$ unit.

(A) The equation of a plane passing through the line of intersection of two planes $P_1: 2x - y = 0$ and $P_2: y - 3z = 0$ is given by $P_1 + \lambda P_2 = 0$.
$(2x - y) + \lambda(y - 3z) = 0$
$2x + (\lambda - 1)y - 3\lambda z = 0$ --- $(i)$
This plane is perpendicular to the plane $4x + 5y - 3z - 8 = 0$.
The normal vectors of these two planes are $\vec{n_1} = 2\hat{i} + (\lambda - 1)\hat{j} - 3\lambda\hat{k}$ and $\vec{n_2} = 4\hat{i} + 5\hat{j} - 3\hat{k}$.
Since the planes are perpendicular,their dot product is zero: $\vec{n_1} \cdot \vec{n_2} = 0$.
$2(4) + (\lambda - 1)(5) + (-3\lambda)(-3) = 0$
$8 + 5\lambda - 5 + 9\lambda = 0$
$14\lambda + 3 = 0$
$\lambda = -\frac{3}{14}$
Substituting $\lambda = -\frac{3}{14}$ into equation $(i)$:
$(2x - y) - \frac{3}{14}(y - 3z) = 0$
$28x - 14y - 3y + 9z = 0$
$28x - 17y + 9z = 0$.

(B) The perpendicular line from a point $(\alpha, \beta, \gamma)$ to a plane $ax + by + cz + d = 0$ passes through the point $(\alpha, \beta, \gamma)$.
Since the line is perpendicular to the plane,its direction ratios are the same as the normal vector of the plane,which are $(a, b, c)$.
The equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.
Substituting the given point,the equation is $\frac{x - \alpha}{a} = \frac{y - \beta}{b} = \frac{z - \gamma}{c}$.

(B) The equation of any plane passing through the line of intersection of the planes $P_1: x + y + z - 1 = 0$ and $P_2: 2x + 3y - z + 4 = 0$ is given by $P_1 + \lambda P_2 = 0$.
$(x + y + z - 1) + \lambda(2x + 3y - z + 4) = 0$
$(1 + 2\lambda)x + (1 + 3\lambda)y + (1 - \lambda)z + (4\lambda - 1) = 0$ ... $(i)$
Since the plane is parallel to the $x$-axis,its normal vector must be perpendicular to the $x$-axis (which has direction vector $\vec{i} = (1, 0, 0)$).
Therefore,the coefficient of $x$ must be zero:
$1 + 2\lambda = 0 \implies \lambda = -\frac{1}{2}$
Substituting $\lambda = -\frac{1}{2}$ into equation $(i)$:
$(1 + 2(-\frac{1}{2}))x + (1 + 3(-\frac{1}{2}))y + (1 - (-\frac{1}{2}))z + (4(-\frac{1}{2}) - 1) = 0$
$0x + (1 - \frac{3}{2})y + (1 + \frac{1}{2})z + (-2 - 1) = 0$
$-\frac{1}{2}y + \frac{3}{2}z - 3 = 0$
Multiplying by $-2$,we get $y - 3z + 6 = 0$.

(D) Let $M$ be the foot of the perpendicular from $P(7, 14, 5)$ to the given plane $2x + 4y - z = 2$. The line $PM$ is normal to the plane,so its direction ratios are $(2, 4, -1)$.
The equation of line $PM$ passing through $(7, 14, 5)$ is $\frac{x - 7}{2} = \frac{y - 14}{4} = \frac{z - 5}{-1} = r$.
Any point on this line is $M(2r + 7, 4r + 14, -r + 5)$.
Since $M$ lies on the plane $2x + 4y - z = 2$,we have:
$2(2r + 7) + 4(4r + 14) - (-r + 5) = 2$
$4r + 14 + 16r + 56 + r - 5 = 2$
$21r + 65 = 2$
$21r = -63$
$r = -3$
Substituting $r = -3$ into the coordinates of $M$:
$x = 2(-3) + 7 = 1$
$y = 4(-3) + 14 = 2$
$z = -(-3) + 5 = 8$
So,the foot of the perpendicular is $M(1, 2, 8)$.
The length of the perpendicular $PM$ is the distance between $(7, 14, 5)$ and $(1, 2, 8)$:
$PM = \sqrt{(1 - 7)^2 + (2 - 14)^2 + (8 - 5)^2}$
$PM = \sqrt{(-6)^2 + (-12)^2 + (3)^2}$
$PM = \sqrt{36 + 144 + 9} = \sqrt{189} = \sqrt{9 \times 21} = 3\sqrt{21}$.
Thus,the length is $3\sqrt{21}$ and the foot is $(1, 2, 8)$.

(A) The equation of any plane passing through the intersection of the planes $P_1: x + y + z - 6 = 0$ and $P_2: 2x + 3y + 4z + 5 = 0$ is given by $P_1 + \lambda P_2 = 0$.
$(x + y + z - 6) + \lambda(2x + 3y + 4z + 5) = 0$.
Since the plane passes through the point $(1, 1, 1)$,we substitute $x = 1, y = 1, z = 1$ into the equation:
$(1 + 1 + 1 - 6) + \lambda(2(1) + 3(1) + 4(1) + 5) = 0$.
$(-3) + \lambda(2 + 3 + 4 + 5) = 0$.
$-3 + 14\lambda = 0 \Rightarrow \lambda = \frac{3}{14}$.
Substituting $\lambda = \frac{3}{14}$ back into the equation:
$(x + y + z - 6) + \frac{3}{14}(2x + 3y + 4z + 5) = 0$.
$14(x + y + z - 6) + 3(2x + 3y + 4z + 5) = 0$.
$14x + 14y + 14z - 84 + 6x + 9y + 12z + 15 = 0$.
$20x + 23y + 26z - 69 = 0$.

(C) Let the plane $x - 2y + 3z - 17 = 0$ divide the line segment joining points $A(-2, 4, 7)$ and $B(3, -5, 8)$ in the ratio $k:1$.
Using the section formula,the coordinates of the point of division are $\left( \frac{3k - 2}{k + 1}, \frac{-5k + 4}{k + 1}, \frac{8k + 7}{k + 1} \right)$.
Since this point lies on the plane $x - 2y + 3z = 17$,we substitute these coordinates into the plane equation:
$\left( \frac{3k - 2}{k + 1} \right) - 2\left( \frac{-5k + 4}{k + 1} \right) + 3\left( \frac{8k + 7}{k + 1} \right) = 17$
Multiplying by $(k + 1)$:
$(3k - 2) - 2(-5k + 4) + 3(8k + 7) = 17(k + 1)$
$3k - 2 + 10k - 8 + 24k + 21 = 17k + 17$
$37k + 11 = 17k + 17$
$20k = 6$
$k = \frac{6}{20} = \frac{3}{10}$
Thus,the required ratio is $3:10$.

(B) The distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by the formula:
$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$
Given the point $(2, 3, -5)$ and the plane $x + 2y - 2z - 9 = 0$,we have $A=1, B=2, C=-2, D=-9$.
Substituting these values into the formula:
$d = \frac{|(1)(2) + (2)(3) + (-2)(-5) - 9|}{\sqrt{1^2 + 2^2 + (-2)^2}}$
$d = \frac{|2 + 6 + 10 - 9|}{\sqrt{1 + 4 + 4}}$
$d = \frac{|9|}{\sqrt{9}}$
$d = \frac{9}{3} = 3$
Thus,the distance is $3$ units.

(B) The equation of a plane passing through the intersection of two planes $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$.
Given planes are $x + 2y + 3z + 4 = 0$ and $4x + 3y + 2z + 1 = 0$.
So,the equation of the required plane is:
$(x + 2y + 3z + 4) + \lambda (4x + 3y + 2z + 1) = 0 \quad .....(i)$
Since this plane passes through the origin $(0, 0, 0)$,we substitute $x = 0, y = 0, z = 0$ into equation $(i)$:
$(0 + 2(0) + 3(0) + 4) + \lambda (4(0) + 3(0) + 2(0) + 1) = 0$
$4 + \lambda(1) = 0$
$\lambda = -4$
Now,substitute $\lambda = -4$ back into equation $(i)$:
$(x + 2y + 3z + 4) - 4(4x + 3y + 2z + 1) = 0$
$x + 2y + 3z + 4 - 16x - 12y - 8z - 4 = 0$
$-15x - 10y - 5z = 0$
Dividing the entire equation by $-5$,we get:
$3x + 2y + z = 0$

(C) The equation of the $XOZ$ plane is $y = 0$.
Let the ratio in which the $XOZ$ plane divides the line segment joining the points $A(x_1, y_1, z_1) = (2, 3, 1)$ and $B(x_2, y_2, z_2) = (6, 7, 1)$ be $k:1$.
The coordinates of the point dividing the segment in the ratio $k:1$ are given by the section formula:
$P = \left( \frac{kx_2 + x_1}{k+1}, \frac{ky_2 + y_1}{k+1}, \frac{kz_2 + z_1}{k+1} \right)$
Since the point $P$ lies on the $XOZ$ plane,its $y$-coordinate must be $0$:
$\frac{ky_2 + y_1}{k+1} = 0$
$k(7) + 3 = 0$
$7k = -3$
$k = -\frac{3}{7}$
Thus,the ratio is $-3:7$.

(A) The equation of any plane passing through the intersection of the planes $P_1: x + y + z - 1 = 0$ and $P_2: 2x + 3y - z + 4 = 0$ is given by $P_1 + \lambda P_2 = 0$.
$(x + y + z - 1) + \lambda (2x + 3y - z + 4) = 0$
Rearranging the terms,we get:
$(1 + 2\lambda)x + (1 + 3\lambda)y + (1 - \lambda)z + (4\lambda - 1) = 0$
Since this plane is parallel to the $x$-axis,its normal vector must be perpendicular to the $x$-axis (which has the direction vector $\vec{i} = (1, 0, 0)$).
Therefore,the coefficient of $x$ must be zero:
$1 + 2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$
Substituting $\lambda = -\frac{1}{2}$ into the equation:
$(1 + 2(-\frac{1}{2}))x + (1 + 3(-\frac{1}{2}))y + (1 - (-\frac{1}{2}))z + (4(-\frac{1}{2}) - 1) = 0$
$0x + (1 - \frac{3}{2})y + (1 + \frac{1}{2})z + (-2 - 1) = 0$
$-\frac{1}{2}y + \frac{3}{2}z - 3 = 0$
Multiplying by $-2$,we get:
$y - 3z + 6 = 0$

(D) The equation of a plane passing through $(0, 1, 2)$ is given by $a(x - 0) + b(y - 1) + c(z - 2) = 0$,which simplifies to $ax + b(y - 1) + c(z - 2) = 0 \dots (i)$.
Since the plane passes through $(-1, 0, 3)$,we substitute these coordinates into $(i)$:
$a(-1 - 0) + b(0 - 1) + c(3 - 2) = 0 \implies -a - b + c = 0 \implies a + b - c = 0 \dots (ii)$.
The plane $(i)$ is perpendicular to the plane $2x + 3y + z = 5$,so the normal vectors are perpendicular,implying $2a + 3b + c = 0 \dots (iii)$.
Solving $(ii)$ and $(iii)$ using cross multiplication:
$\frac{a}{1(1) - (-1)(3)} = \frac{-b}{1(1) - (-1)(2)} = \frac{c}{1(3) - 1(2)}$
$\frac{a}{1 + 3} = \frac{-b}{1 + 2} = \frac{c}{3 - 2} \implies \frac{a}{4} = \frac{-b}{3} = \frac{c}{1} = k$.
Thus,$a = 4k, b = -3k, c = k$.
Substituting these into $(i)$:
$4k(x) - 3k(y - 1) + k(z - 2) = 0$.
Dividing by $k$ $(k \neq 0)$:
$4x - 3y + 3 + z - 2 = 0 \implies 4x - 3y + z + 1 = 0$.

(B) Let the line joining the points $A(2, -3, 1)$ and $B(3, -4, -5)$ be divided by the plane $2x + y + z = 7$ in the ratio $k:1$.
The coordinates of the point of intersection $P$ are given by the section formula:
$P = \left( \frac{3k + 2}{k + 1}, \frac{-4k - 3}{k + 1}, \frac{-5k + 1}{k + 1} \right)$
Since $P$ lies on the plane $2x + y + z = 7$,we substitute these coordinates into the plane equation:
$2\left( \frac{3k + 2}{k + 1} \right) + \left( \frac{-4k - 3}{k + 1} \right) + \left( \frac{-5k + 1}{k + 1} \right) = 7$
Multiply by $(k + 1)$:
$2(3k + 2) + (-4k - 3) + (-5k + 1) = 7(k + 1)$
$6k + 4 - 4k - 3 - 5k + 1 = 7k + 7$
$-3k + 2 = 7k + 7$
$-10k = 5 \implies k = -\frac{1}{2}$
Substituting $k = -\frac{1}{2}$ back into the coordinates of $P$:
$x = \frac{3(-1/2) + 2}{-1/2 + 1} = \frac{-1.5 + 2}{0.5} = \frac{0.5}{0.5} = 1$
$y = \frac{-4(-1/2) - 3}{-1/2 + 1} = \frac{2 - 3}{0.5} = \frac{-1}{0.5} = -2$
$z = \frac{-5(-1/2) + 1}{-1/2 + 1} = \frac{2.5 + 1}{0.5} = \frac{3.5}{0.5} = 7$
Thus,the point of intersection is $(1, -2, 7)$.

(D) Let the given line be $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3} = r$.
Then,the coordinates of any point on the line are given by $x = r$,$y = 2r + 1$,and $z = 3r - 2$.
Since this point lies on the plane $2x + 3y + z = 0$,we substitute these values into the plane equation:
$2(r) + 3(2r + 1) + (3r - 2) = 0$.
Expanding the equation:
$2r + 6r + 3 + 3r - 2 = 0$.
Combining like terms:
$11r + 1 = 0$,which gives $r = \frac{-1}{11}$.
Now,substitute $r = \frac{-1}{11}$ back into the expressions for $x, y, z$:
$x = \frac{-1}{11}$,
$y = 2(\frac{-1}{11}) + 1 = \frac{-2}{11} + \frac{11}{11} = \frac{9}{11}$,
$z = 3(\frac{-1}{11}) - 2 = \frac{-3}{11} - \frac{22}{11} = \frac{-25}{11}$.
Thus,the point of intersection is $(\frac{-1}{11}, \frac{9}{11}, \frac{-25}{11})$.

(A) The direction ratios of the given line are $(3, 4, 0)$.
The normal vector to the $xy$-plane is $\vec{n} = (0, 0, 1)$.
$A$ line with direction vector $\vec{b} = (a, b, c)$ is parallel to a plane with normal $\vec{n} = (n_1, n_2, n_3)$ if $\vec{b} \cdot \vec{n} = 0$.
Here,$(3, 4, 0) \cdot (0, 0, 1) = (3 \times 0) + (4 \times 0) + (0 \times 1) = 0$.
Since the dot product is $0$,the line is parallel to the $xy$-plane.

(A) The given equation of the line is $x = 2y = 3z$.
Dividing by $6$,we get $\frac{x}{6} = \frac{2y}{6} = \frac{3z}{6}$,which simplifies to $\frac{x}{6} = \frac{y}{3} = \frac{z}{2}$.
The direction ratios of the line are $(6, 3, 2)$.
Since the plane is perpendicular to this line,the direction ratios of the line are the normal vector components $(a, b, c)$ of the plane.
The equation of a plane passing through the origin $(0, 0, 0)$ is $a(x - 0) + b(y - 0) + c(z - 0) = 0$,where $(a, b, c)$ are the direction ratios of the normal.
Substituting $(a, b, c) = (6, 3, 2)$,we get $6x + 3y + 2z = 0$.

(A) The direction ratios of the line are $(a_1, b_1, c_1) = (2, 3, 2)$.
The normal vector to the plane $4x - 2y - z = 1$ has direction ratios $(a_2, b_2, c_2) = (4, -2, -1)$.
For a line to be parallel to a plane,the dot product of the direction ratios of the line and the normal vector of the plane must be zero,i.e.,$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.
Calculating the dot product: $(2)(4) + (3)(-2) + (2)(-1) = 8 - 6 - 2 = 0$.
Since the dot product is $0$,the line is parallel to the plane.
Next,we check if the line lies in the plane by testing a point on the line $(-3, 4, -5)$ in the plane equation: $4(-3) - 2(4) - (-5) = -12 - 8 + 5 = -15 \neq 1$.
Since the point does not satisfy the plane equation,the line is strictly parallel to the plane.

(A) The equation of a line passing through the point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.
The given point is $(1, 2, 3)$,so the equation is $\frac{x - 1}{a} = \frac{y - 2}{b} = \frac{z - 3}{c}$.
$A$ line is perpendicular to a plane if its direction ratios are proportional to the normal vector of the plane.
The equation of the plane is $x + 2y - 5z + 9 = 0$,so the normal vector is $\vec{n} = (1, 2, -5)$.
Thus,the direction ratios of the line are $(1, 2, -5)$.
Substituting these into the line equation,we get $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{-5}$.

(D) Let the equation of the plane be $a(x - 4) + b(y - 3) + c(z - 2) = 0$.
Since the plane contains the lines,the normal vector $(a, b, c)$ is perpendicular to the direction vectors of both lines,$(1, 1, 2)$ and $(1, -4, 5)$.
Thus,$a + b + 2c = 0$ and $a - 4b + 5c = 0$.
Using the cross product to find the normal vector $(a, b, c) = (1, 1, 2) \times (1, -4, 5)$:
$a = (1)(5) - (2)(-4) = 5 + 8 = 13$.
$b = (2)(1) - (1)(5) = 2 - 5 = -3$.
$c = (1)(-4) - (1)(1) = -4 - 1 = -5$.
So the normal vector is $(13, -3, -5)$.
The equation of the plane is $13(x - 4) - 3(y - 3) - 5(z - 2) = 0$.
$13x - 52 - 3y + 9 - 5z + 10 = 0$.
$13x - 3y - 5z - 33 = 0$ or $13x - 3y - 5z = 33$.
Since this does not match any of the given options,the correct answer is $(d)$.

(D) The equation of a plane passing through $(1, 0, -1)$ is given by $a(x - 1) + b(y - 0) + c(z + 1) = 0 \dots (i)$.
Since the plane passes through $(3, 2, 2)$,we have $a(3 - 1) + b(2 - 0) + c(2 + 1) = 0$,which simplifies to $2a + 2b + 3c = 0 \dots (ii)$.
The plane is parallel to the line with direction ratios $(2, -2, 3)$,so the normal to the plane is perpendicular to the line. Thus,$2a - 2b + 3c = 0 \dots (iii)$.
Subtracting $(iii)$ from $(ii)$,we get $4b = 0$,so $b = 0$.
Substituting $b = 0$ into $(ii)$,we get $2a + 3c = 0$,or $a = -\frac{3}{2}c$. Let $c = 2$,then $a = -3$.
The normal vector is $(-3, 0, 2)$.
The equation of the plane is $-3(x - 1) + 0(y - 0) + 2(z + 1) = 0$,which simplifies to $-3x + 3 + 2z + 2 = 0$,or $3x - 2z - 5 = 0$.

(B) line with direction ratios $(l, m, n)$ is parallel to a plane $ax + by + cz + d = 0$ if and only if $al + bm + cn = 0$ and the line does not lie on the plane.
Given the line $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$,the direction ratios are $(l, m, n) = (3, 4, 5)$.
Checking option $(b)$: $2x + y - 2z = 0$. Here $(a, b, c) = (2, 1, -2)$.
Calculating $al + bm + cn = 2(3) + 1(4) + (-2)(5) = 6 + 4 - 10 = 0$.
Since the condition $al + bm + cn = 0$ is satisfied,the line is parallel to the plane $2x + y - 2z = 0$.

(D) Let any point on the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12} = t$ be $(3t + 2, 4t - 1, 12t + 2)$.
This point lies on the plane $x - y + z = 5$.
Substituting the coordinates into the plane equation:
$(3t + 2) - (4t - 1) + (12t + 2) = 5$
$3t + 2 - 4t + 1 + 12t + 2 = 5$
$11t + 5 = 5$
$11t = 0 \Rightarrow t = 0$.
Thus,the point of intersection is $(3(0) + 2, 4(0) - 1, 12(0) + 2) = (2, -1, 2)$.
The distance between $(2, -1, 2)$ and $(-1, -5, -10)$ is given by the distance formula:
$d = \sqrt{(-1 - 2)^2 + (-5 - (-1))^2 + (-10 - 2)^2}$
$d = \sqrt{(-3)^2 + (-4)^2 + (-12)^2}$
$d = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$.

(A) Let the direction ratios of the line be $(l, m, n)$.
Since the line is parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$,it must be perpendicular to the normals of these planes.
The normals are $\vec{n_1} = (1, -1, 2)$ and $\vec{n_2} = (3, 1, 1)$.
Thus,$l - m + 2n = 0$ and $3l + m + n = 0$.
Using the cross product to find the direction ratios $(l, m, n) = \vec{n_1} \times \vec{n_2}$:
$l = (-1)(1) - (2)(1) = -3$
$m = (2)(3) - (1)(1) = 5$
$n = (1)(1) - (-1)(3) = 4$
So,the direction ratios are $(-3, 5, 4)$.
The line passes through $(1, 2, 3)$,so its equation is $\frac{x - 1}{-3} = \frac{y - 2}{5} = \frac{z - 3}{4}$.

(B) Let the line be $\frac{x + 3}{3} = \frac{y - 2}{-2} = \frac{z + 1}{1} = \lambda$.
Then,$x = 3\lambda - 3$,$y = -2\lambda + 2$,and $z = \lambda - 1$.
Since the line intersects the plane $4x + 5y + 3z - 5 = 0$,the point $(3\lambda - 3, -2\lambda + 2, \lambda - 1)$ must satisfy the plane equation.
Substituting these values into the plane equation:
$4(3\lambda - 3) + 5(-2\lambda + 2) + 3(\lambda - 1) - 5 = 0$
$12\lambda - 12 - 10\lambda + 10 + 3\lambda - 3 - 5 = 0$
$5\lambda - 10 = 0$
$5\lambda = 10 \Rightarrow \lambda = 2$.
Substituting $\lambda = 2$ back into the expressions for $x, y, z$:
$x = 3(2) - 3 = 3$
$y = -2(2) + 2 = -2$
$z = 2 - 1 = 1$.
Thus,the point of intersection is $(3, -2, 1)$.

(B) The direction ratios of the line are $(l, m, n)$.
The normal vector to the plane $ax + by + cz + d = 0$ is $\vec{n} = (a, b, c)$.
If a line is parallel to a plane,the line is perpendicular to the normal of the plane.
Therefore,the dot product of the direction vector of the line and the normal vector of the plane must be zero.
$(l, m, n) \cdot (a, b, c) = 0$
$al + bm + cn = 0$.

(C) The equation of a plane passing through the line of intersection of the planes $ax + by + cz + d = 0$ and $a'x + b'y + c'z + d' = 0$ is given by $(ax + by + cz + d) + \lambda (a'x + b'y + c'z + d') = 0$.
Rearranging the terms,we get $x(a + \lambda a') + y(b + \lambda b') + z(c + \lambda c') + (d + \lambda d') = 0$ ... $(i)$.
Since this plane is parallel to the line $y = 0, z = 0$ (which is the $x$-axis),its normal vector must be perpendicular to the direction vector of the $x$-axis,which is $(1, 0, 0)$.
Therefore,the dot product of the normal vector $(a + \lambda a', b + \lambda b', c + \lambda c')$ and $(1, 0, 0)$ must be zero:
$1(a + \lambda a') + 0(b + \lambda b') + 0(c + \lambda c') = 0$.
This gives $a + \lambda a' = 0$,so $\lambda = -\frac{a}{a'}$.
Substituting $\lambda = -\frac{a}{a'}$ into equation $(i)$:
$(ax + by + cz + d) - \frac{a}{a'}(a'x + b'y + c'z + d') = 0$.
Multiplying by $a'$,we get $a'ax + a'by + a'cz + a'd - aa'x - ab'y - ac'z - ad' = 0$.
Simplifying,we get $(a'b - ab')y + (a'c - ac')z + (a'd - ad') = 0$.
Multiplying by $-1$,we obtain $(ab' - a'b)y + (ac' - a'c)z + (ad' - a'd) = 0$.

(B) line with direction ratios $(a, b, c)$ is parallel to a plane with normal vector $(A, B, C)$ if the dot product of the direction vector of the line and the normal vector of the plane is zero,i.e.,$aA + bB + cC = 0$.
The direction ratios of the given line are $(3, 4, 5)$.
Checking the options:
$(a)$ Normal vector is $(2, 3, 4)$. Dot product: $(3)(2) + (4)(3) + (5)(4) = 6 + 12 + 20 = 38 \neq 0$.
$(b)$ Normal vector is $(3, 4, -5)$. Dot product: $(3)(3) + (4)(4) + (5)(-5) = 9 + 16 - 25 = 0$.
$(c)$ Normal vector is $(3, 4, 5)$. Dot product: $(3)(3) + (4)(4) + (5)(5) = 9 + 16 + 25 = 50 \neq 0$.
$(d)$ Normal vector is $(1, 1, 1)$. Dot product: $(3)(1) + (4)(1) + (5)(1) = 12 \neq 0$.
Since the dot product is zero for option $(b)$,the line is parallel to the plane $3x + 4y - 5z = 10$.

(D) First,check if the line is parallel to the plane. The direction vector of the line is $\vec{v} = (3, -2, 2)$ and the normal to the plane is $\vec{n} = (2, 2, -1)$.
Since $\vec{v} \cdot \vec{n} = (3)(2) + (-2)(2) + (2)(-1) = 6 - 4 - 2 = 0$,the line is parallel to the plane.
To find the distance,pick any point on the line,such as $P(1, -2, 1)$.
The perpendicular distance $d$ from point $(x_1, y_1, z_1)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.
Substituting the values: $d = \frac{|2(1) + 2(-2) - 1(1) - 6|}{\sqrt{2^2 + 2^2 + (-1)^2}} = \frac{|2 - 4 - 1 - 6|}{\sqrt{4 + 4 + 1}} = \frac{|-9|}{\sqrt{9}} = \frac{9}{3} = 3$.

(C) The line passes through the point $P(1, 2, 3)$ and has direction ratios $(5, 4, 5)$.
Since the plane passes through the origin $(0, 0, 0)$ and the point $P(1, 2, 3)$,the vector $\vec{OP} = \hat{i} + 2\hat{j} + 3\hat{k}$ lies in the plane.
The direction vector of the line is $\vec{v} = 5\hat{i} + 4\hat{j} + 5\hat{k}$.
The normal to the plane $\vec{n}$ is given by the cross product $\vec{OP} \times \vec{v}$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 5 & 4 & 5 \end{vmatrix} = \hat{i}(10 - 12) - \hat{j}(5 - 15) + \hat{k}(4 - 10) = -2\hat{i} + 10\hat{j} - 6\hat{k}$.
Dividing by $-2$,we get the normal vector $\vec{n}' = \hat{i} - 5\hat{j} + 3\hat{k}$.
The equation of the plane passing through $(0, 0, 0)$ is $1(x - 0) - 5(y - 0) + 3(z - 0) = 0$,which simplifies to $x - 5y + 3z = 0$.

(D) The direction ratios of the line are $\vec{v} = (a, b, c)$.
The normal vector to the plane $ax + by + cz + 6 = 0$ is $\vec{n} = (a, b, c)$.
Since the direction ratios of the line are proportional to the direction ratios of the normal to the plane,the line is perpendicular to the plane.
The angle $\theta$ between a line with direction vector $\vec{v}$ and a plane with normal vector $\vec{n}$ is given by $\sin \theta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|}$.
Substituting the values,$\sin \theta = \frac{|a^2 + b^2 + c^2|}{\sqrt{a^2 + b^2 + c^2} \sqrt{a^2 + b^2 + c^2}} = \frac{a^2 + b^2 + c^2}{a^2 + b^2 + c^2} = 1$.
Therefore,$\sin \theta = 1$,which implies $\theta = 90^o$.

(D) Let the point on the line be $(x, y, z)$.
Given the line equation: $\frac{x - 6}{-1} = \frac{y + 1}{0} = \frac{z + 3}{4} = r$.
Expressing coordinates in terms of $r$: $x = -r + 6$,$y = -1$,$z = 4r - 3$.
Since the point lies on the plane $x + y - z = 3$,substitute these values into the plane equation:
$(-r + 6) + (-1) - (4r - 3) = 3$.
$-r + 6 - 1 - 4r + 3 = 3$.
$-5r + 8 = 3$.
$-5r = -5$.
$r = 1$.
Substituting $r = 1$ back into the coordinate expressions:
$x = -1 + 6 = 5$.
$y = -1$.
$z = 4(1) - 3 = 1$.
Thus,the required coordinates are $(5, -1, 1)$.

(C) The direction vector of the line is $\vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}$.
The normal vector to the plane $x + y + 4 = 0$ is $\vec{n} = 1\hat{i} + 1\hat{j} + 0\hat{k}$.
The angle $\theta$ between a line with direction vector $\vec{b}$ and a plane with normal $\vec{n}$ is given by $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
Calculating the dot product: $\vec{b} \cdot \vec{n} = (2)(1) + (1)(1) + (-2)(0) = 2 + 1 + 0 = 3$.
Calculating the magnitudes: $|\vec{b}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$ and $|\vec{n}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$.
Substituting these values: $\sin \theta = \frac{|3|}{3 \times \sqrt{2}} = \frac{1}{\sqrt{2}}$.
Therefore,$\theta = \arcsin\left(\frac{1}{\sqrt{2}}\right) = 45^o$.