Line Questions in English

(B) Let the vertices be $A(3, 2, 0)$,$B(5, 3, 2)$,and $C(-9, 6, -3)$.
First,calculate the lengths of sides $AB$ and $AC$:
$AB = \sqrt{(5-3)^2 + (3-2)^2 + (2-0)^2} = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.
$AC = \sqrt{(-9-3)^2 + (6-2)^2 + (-3-0)^2} = \sqrt{(-12)^2 + 4^2 + (-3)^2} = \sqrt{144 + 16 + 9} = \sqrt{169} = 13$.
According to the Angle Bisector Theorem,the bisector $AD$ divides the side $BC$ in the ratio $AB : AC = 3 : 13$.
Using the section formula,the coordinates of $D$ are given by:
$D = \left( \frac{3(-9) + 13(5)}{3+13}, \frac{3(6) + 13(3)}{3+13}, \frac{3(-3) + 13(2)}{3+13} \right)$
$D = \left( \frac{-27 + 65}{16}, \frac{18 + 39}{16}, \frac{-9 + 26}{16} \right)$
$D = \left( \frac{38}{16}, \frac{57}{16}, \frac{17}{16} \right) = \left( \frac{19}{8}, \frac{57}{16}, \frac{17}{16} \right)$.
Thus,option $B$ is correct.

(B) The lines are given by $\vec{r_1} = (6, 2, 2) + \lambda(1, -2, 2)$ and $\vec{r_2} = (-4, 0, -1) + \mu(3, -2, -2)$.
Let $\vec{a_1} = (6, 2, 2)$,$\vec{a_2} = (-4, 0, -1)$,$\vec{b_1} = (1, -2, 2)$,and $\vec{b_2} = (3, -2, -2)$.
The shortest distance is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{ |\vec{b_1} \times \vec{b_2}| }$.
First,calculate $\vec{a_2} - \vec{a_1} = (-4-6, 0-2, -1-2) = (-10, -2, -3)$.
Next,calculate $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{vmatrix} = \hat{i}(4 - (-4)) - \hat{j}(-2 - 6) + \hat{k}(-2 - (-6)) = 8\hat{i} + 8\hat{j} + 4\hat{k}$.
The magnitude $|\vec{b_1} \times \vec{b_2}| = \sqrt{8^2 + 8^2 + 4^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12$.
The dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-10)(8) + (-2)(8) + (-3)(4) = -80 - 16 - 12 = -108$.
Thus,$d = \frac{|-108|}{12} = \frac{108}{12} = 9$.

(C) Let the point be $P(2, -1, 5)$ and the line be $L: \frac{x - 11}{10} = \frac{y + 2}{-4} = \frac{z + 8}{-11} = k$.
Any point $Q$ on the line is given by $(10k + 11, -4k - 2, -11k - 8)$.
The direction ratios of the line $PQ$ are $(10k + 11 - 2, -4k - 2 + 1, -11k - 8 - 5) = (10k + 9, -4k - 1, -11k - 13)$.
Since $PQ$ is perpendicular to the line $L$ with direction ratios $(10, -4, -11)$,their dot product must be zero:
$10(10k + 9) - 4(-4k - 1) - 11(-11k - 13) = 0$.
$100k + 90 + 16k + 4 + 121k + 143 = 0$.
$237k + 237 = 0 \implies k = -1$.
Substituting $k = -1$ into the coordinates of $Q$:
$x = 10(-1) + 11 = 1$,
$y = -4(-1) - 2 = 2$,
$z = -11(-1) - 8 = 3$.
So,the foot of the perpendicular is $(1, 2, 3)$.
The length of the perpendicular $PQ$ is $\sqrt{(1 - 2)^2 + (2 - (-1))^2 + (3 - 5)^2} = \sqrt{(-1)^2 + 3^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$.

(B) The given lines can be written in symmetric form as follows:
For the first line $x = ay + b$ and $z = cy + d$,we have $\frac{x - b}{a} = y = \frac{z - d}{c}$.
Thus,the direction ratios of the first line are $(a, 1, c)$.
For the second line $x = a'y + b'$ and $z = c'y + d'$,we have $\frac{x - b'}{a'} = y = \frac{z - d'}{c'}$.
Thus,the direction ratios of the second line are $(a', 1, c')$.
Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.
Substituting the direction ratios,we get $(a)(a') + (1)(1) + (c)(c') = 0$.
Therefore,$aa' + 1 + cc' = 0$,which implies $aa' + cc' = -1$.

(C) Let the first line be $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} = \lambda$.
Any point on this line is $(3\lambda - 1, 5\lambda - 3, 7\lambda - 5)$.
Let the second line be $\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5} = \mu$.
Any point on this line is $(\mu + 2, 3\mu + 4, 5\mu + 6)$.
For the lines to intersect,the coordinates must be equal:
$3\lambda - 1 = \mu + 2 \implies 3\lambda - \mu = 3$
$5\lambda - 3 = 3\mu + 4 \implies 5\lambda - 3\mu = 7$
Solving these equations:
Multiply the first by $3$: $9\lambda - 3\mu = 9$.
Subtract the second from this: $(9\lambda - 3\mu) - (5\lambda - 3\mu) = 9 - 7 \implies 4\lambda = 2 \implies \lambda = \frac{1}{2}$.
Substituting $\lambda = \frac{1}{2}$ into $3\lambda - \mu = 3$:
$3(\frac{1}{2}) - \mu = 3 \implies \mu = \frac{3}{2} - 3 = -\frac{3}{2}$.
Now,find the point using $\lambda = \frac{1}{2}$:
$x = 3(\frac{1}{2}) - 1 = \frac{1}{2}$,
$y = 5(\frac{1}{2}) - 3 = -\frac{1}{2}$,
$z = 7(\frac{1}{2}) - 5 = -\frac{3}{2}$.
The point of intersection is $(\frac{1}{2}, -\frac{1}{2}, -\frac{3}{2})$.

(B) To find the symmetrical form of the line of intersection of two planes $P_1: 3x + 2y - z - 4 = 0$ and $P_2: 4x + y - 2z + 3 = 0$,we first find the direction vector $\vec{v}$ of the line,which is the cross product of the normals $\vec{n_1} = (3, 2, -1)$ and $\vec{n_2} = (4, 1, -2)$.
$\vec{v} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -1 \\ 4 & 1 & -2 \end{vmatrix} = \hat{i}(-4 + 1) - \hat{j}(-6 + 4) + \hat{k}(3 - 8) = -3\hat{i} + 2\hat{j} - 5\hat{k}$.
We can take the direction ratios as $(3, -2, 5)$.
Now,find a point on the line by setting $z = 0$ in the plane equations:
$3x + 2y = 4$ and $4x + y = -3$.
Multiplying the second equation by $2$: $8x + 2y = -6$.
Subtracting the first from this: $5x = -10 \implies x = -2$.
Substituting $x = -2$ into $4x + y = -3$: $4(-2) + y = -3 \implies y = 5$.
So,the point is $(-2, 5, 0)$.
The symmetrical form is $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$,which gives $\frac{x + 2}{3} = \frac{y - 5}{-2} = \frac{z}{5}$.

(B) The lines are given by $\frac{x - 3}{2} = \frac{y + 1}{-3} = \frac{z + a}{p}$ and $\frac{x + 2}{2} = \frac{y - 4}{4} = \frac{z + 5}{2}$.
Since the lines are perpendicular,the dot product of their direction vectors $(2, -3, p)$ and $(2, 4, 2)$ must be zero:
$(2)(2) + (-3)(4) + (p)(2) = 0$
$4 - 12 + 2p = 0 \Rightarrow 2p = 8 \Rightarrow p = 4$.
Since the lines are coplanar,the determinant of the vector connecting points $(3, -1, -a)$ and $(-2, 4, -5)$ and the two direction vectors must be zero:
$\left|\begin{array}{ccc} 3 - (-2) & -1 - 4 & -a - (-5) \\ 2 & -3 & p \\ 2 & 4 & 2 \end{array}\right| = 0$
$\left|\begin{array}{ccc} 5 & -5 & 5 - a \\ 2 & -3 & 4 \\ 2 & 4 & 2 \end{array}\right| = 0$
$5(-6 - 16) + 5(4 - 8) + (5 - a)(8 + 6) = 0$
$5(-22) + 5(-4) + 14(5 - a) = 0$
$-110 - 20 + 70 - 14a = 0$
$-60 - 14a = 0 \Rightarrow 14a = -60 \Rightarrow a = -\frac{30}{7}$.
Thus,$a + p = -\frac{30}{7} + 4 = \frac{-30 + 28}{7} = -\frac{2}{7}$.

(C) The lines are given by $\vec{r} = \vec{a} + \lambda \vec{p}$ and $\vec{r} = \vec{b} + \mu \vec{q}$.
Here,$\vec{a} = 2\hat{i} + 4\hat{j} + 5\hat{k}$ and $\vec{p} = 3\hat{i} + 4\hat{j} + 5\hat{k}$.
Also,$\vec{b} = 1\hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{q} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
The shortest distance $SD$ is given by the formula $SD = \left| \frac{(\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right|$.
First,calculate $\vec{b} - \vec{a} = (1-2)\hat{i} + (2-4)\hat{j} + (3-5)\hat{k} = -\hat{i} - 2\hat{j} - 2\hat{k}$.
Next,calculate $\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 2 & 3 & 4 \end{vmatrix} = \hat{i}(16-15) - \hat{j}(12-10) + \hat{k}(9-8) = \hat{i} - 2\hat{j} + \hat{k}$.
The magnitude $|\vec{p} \times \vec{q}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1+4+1} = \sqrt{6}$.
The dot product $(\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q}) = (-1)(1) + (-2)(-2) + (-2)(1) = -1 + 4 - 2 = 1$.
Thus,$SD = \left| \frac{1}{\sqrt{6}} \right| = \frac{1}{\sqrt{6}}$.

(B) The direction ratios $(l_1, m_1, n_1)$ of the first line are given by the cross product of the normals to the planes $3x + 2y + z = 0$ and $x + y - 2z = 0$.
Using the cross product formula:
$l_1 = (2)(-2) - (1)(1) = -5$
$m_1 = (1)(1) - (3)(-2) = 7$
$n_1 = (3)(1) - (2)(1) = 1$
So,the direction ratios are proportional to $(-5, 7, 1)$ or $(5, -7, -1)$.
The direction ratios $(l_2, m_2, n_2)$ of the second line are given by the cross product of the normals to the planes $2x - y - z = 0$ and $7x + 10y - 8z = 0$.
$l_2 = (-1)(-8) - (-1)(10) = 8 + 10 = 18$
$m_2 = (-1)(7) - (2)(-8) = -7 + 16 = 9$
$n_2 = (2)(10) - (-1)(7) = 20 + 7 = 27$
Dividing by $9$,we get the direction ratios as $(2, 1, 3)$.
Let $\theta$ be the angle between the two lines. The cosine of the angle is given by:
$\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$
$\cos \theta = \frac{|(5)(2) + (-7)(1) + (-1)(3)|}{\sqrt{5^2 + (-7)^2 + (-1)^2} \sqrt{2^2 + 1^2 + 3^2}}$
$\cos \theta = \frac{|10 - 7 - 3|}{\sqrt{25 + 49 + 1} \sqrt{4 + 1 + 9}} = \frac{0}{\sqrt{75} \sqrt{14}} = 0$
Since $\cos \theta = 0$,the angle $\theta = \frac{\pi}{2}$.

(D) The given lines are $L_1: \frac{x - 3}{3} = \frac{y - 8}{-1} = \frac{z - 3}{1}$ and $L_2: \frac{x + 3}{-3} = \frac{y + 7}{2} = \frac{z - 6}{4}$.
The points on the lines are $A(3, 8, 3)$ and $B(-3, -7, 6)$.
The direction vectors are $\vec{b_1} = 3\hat{i} - \hat{j} + \hat{k}$ and $\vec{b_2} = -3\hat{i} + 2\hat{j} + 4\hat{k}$.
Calculate the cross product $\vec{b_1} \times \vec{b_2}$:
$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} = \hat{i}(-4 - 2) - \hat{j}(12 - (-3)) + \hat{k}(6 - 3) = -6\hat{i} - 15\hat{j} + 3\hat{k}$.
The vector $\vec{AB} = (-3 - 3)\hat{i} + (-7 - 8)\hat{j} + (6 - 3)\hat{k} = -6\hat{i} - 15\hat{j} + 3\hat{k}$.
The shortest distance $d$ is given by $d = \frac{|\vec{AB} \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$.
$\vec{AB} \cdot (\vec{b_1} \times \vec{b_2}) = (-6)(-6) + (-15)(-15) + (3)(3) = 36 + 225 + 9 = 270$.
$|\vec{b_1} \times \vec{b_2}| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}$.
$d = \frac{270}{3\sqrt{30}} = \frac{90}{\sqrt{30}} = \frac{90\sqrt{30}}{30} = 3\sqrt{30}$.

(B) For Statement-$1$: The lines are $L_1: \vec{r} = (-3, 6, 0) + \lambda(-4, 3, 2)$ and $L_2: \vec{r} = (-3, 0, 7) + \mu(-4, 1, 1)$.
The shortest distance $d$ between two skew lines $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$ is given by $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$.
Here,$\vec{a}_2 - \vec{a}_1 = (-3 - (-3), 0 - 6, 7 - 0) = (0, -6, 7)$.
$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 3 & 2 \\ -4 & 1 & 1 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(-4+8) + \hat{k}(-4+12) = (1, -4, 8)$.
$|\vec{b}_1 \times \vec{b}_2| = \sqrt{1^2 + (-4)^2 + 8^2} = \sqrt{1 + 16 + 64} = \sqrt{81} = 9$.
$|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)| = |(0, -6, 7) \cdot (1, -4, 8)| = |0 + 24 + 56| = 80$.
So,$d = \frac{80}{9} \neq 9$. Thus,Statement-$1$ is false.
For Statement-$2$: By definition,skew lines are lines in three-dimensional space that are neither parallel nor intersecting,meaning they do not lie in the same plane. Thus,Statement-$2$ is true.

(C) The given equations represent a line in space,which implies that the system of equations must have infinitely many solutions.
The system is given by:
$2x - y + z = 0$
$-x + y + 2z = 0$
$mx - 2y + mz = 0$
For the system to have a non-trivial solution (representing a line passing through the origin),the determinant of the coefficient matrix must be zero:
$\left|\begin{array}{ccc} 2 & -1 & 1 \\ -1 & 1 & 2 \\ m & -2 & m \end{array}\right| = 0$
Expanding the determinant along the first row:
$2(1(m) - 2(-2)) - (-1)((-1)(m) - 2(m)) + 1((-1)(-2) - 1(m)) = 0$
$2(m + 4) + 1(-m - 2m) + 1(2 - m) = 0$
$2m + 8 - 3m + 2 - m = 0$
$-2m + 10 = 0$
$2m = 10$
$m = 5$

(D) The points $A$ and $B$ represent two lines in $3D$ space.
Line $L_1$ passing through $A$ can be written as $\vec{r} = (2, 1, 2) + \lambda(1, -2, 1)$.
Line $L_2$ passing through $B$ can be written as $\vec{r} = (1, 0, 1) + k(2, 1, 1)$.
The shortest distance between two skew lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + k \vec{b_2}$ is given by $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$.
Here,$\vec{a_2} - \vec{a_1} = (1-2, 0-1, 1-2) = (-1, -1, -1)$.
$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 2 & 1 & 1 \end{vmatrix} = \hat{i}(-2-1) - \hat{j}(1-2) + \hat{k}(1+4) = -3\hat{i} + \hat{j} + 5\hat{k}$.
Magnitude $|\vec{b_1} \times \vec{b_2}| = \sqrt{(-3)^2 + 1^2 + 5^2} = \sqrt{9 + 1 + 25} = \sqrt{35}$.
Shortest distance $d = \left| \frac{(-1, -1, -1) \cdot (-3, 1, 5)}{\sqrt{35}} \right| = \left| \frac{3 - 1 - 5}{\sqrt{35}} \right| = \left| \frac{-3}{\sqrt{35}} \right| = \frac{3}{\sqrt{35}}$.

(B) The equation of the line is $\frac{x}{1} = \frac{y}{1} = \frac{z}{-1} = \lambda$.
Any point on this line is of the form $A(\lambda, \lambda, -\lambda)$.
Given that the distance $PA = \sqrt{3}$,we have $PA^2 = 3$.
$(\lambda - 1)^2 + (\lambda - 1)^2 + (-\lambda - 1)^2 = 3$.
$(\lambda^2 - 2\lambda + 1) + (\lambda^2 - 2\lambda + 1) + (\lambda^2 + 2\lambda + 1) = 3$.
$3\lambda^2 - 2\lambda + 3 = 3$.
$3\lambda^2 - 2\lambda = 0$.
$\lambda(3\lambda - 2) = 0$.
Thus,$\lambda = 0$ or $\lambda = \frac{2}{3}$.
The points are $A(0, 0, 0)$ and $B(\frac{2}{3}, \frac{2}{3}, -\frac{2}{3})$.
The distance $AB = \sqrt{(\frac{2}{3} - 0)^2 + (\frac{2}{3} - 0)^2 + (-\frac{2}{3} - 0)^2}$.
$AB = \sqrt{\frac{4}{9} + \frac{4}{9} + \frac{4}{9}} = \sqrt{\frac{12}{9}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.

(A) The lines are given by $L_1: \bar{r} = \bar{a} + \lambda \bar{b}$ and $L_2: \bar{r} = \bar{c} + \mu \bar{d}$.
Here,$\bar{a} = \hat{i} + \hat{j}$,$\bar{b} = \hat{i} + \hat{j} - \hat{k}$,$\bar{c} = \hat{j} + \hat{k}$,and $\bar{d} = -\hat{i} + \hat{j} + 2\hat{k}$.
First,calculate $\bar{a} - \bar{c} = (\hat{i} + \hat{j}) - (\hat{j} + \hat{k}) = \hat{i} - \hat{k}$.
Next,calculate the cross product $\bar{b} \times \bar{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ -1 & 1 & 2 \end{vmatrix} = \hat{i}(2 - (-1)) - \hat{j}(2 - 1) + \hat{k}(1 - (-1)) = 3\hat{i} - \hat{j} + 2\hat{k}$.
The magnitude $|\bar{b} \times \bar{d}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$.
The shortest distance $D$ is given by $D = \frac{|(\bar{a} - \bar{c}) \cdot (\bar{b} \times \bar{d})|}{|\bar{b} \times \bar{d}|}$.
$D = \frac{|(\hat{i} - \hat{k}) \cdot (3\hat{i} - \hat{j} + 2\hat{k})|}{\sqrt{14}} = \frac{|(1)(3) + (0)(-1) + (-1)(2)|}{\sqrt{14}} = \frac{|3 - 2|}{\sqrt{14}} = \frac{1}{\sqrt{14}}$.

(B) Let $D$ be the midpoint of $BC$. The coordinates of $D$ are $\left( \frac{\lambda - 1}{2}, \frac{5 + 3}{2}, \frac{\mu + 2}{2} \right) = \left( \frac{\lambda - 1}{2}, 4, \frac{\mu + 2}{2} \right)$.
The direction ratios of the median $AD$ are $\left( \frac{\lambda - 1}{2} - 2, 4 - 3, \frac{\mu + 2}{2} - 5 \right) = \left( \frac{\lambda - 5}{2}, 1, \frac{\mu - 8}{2} \right)$.
Since the median $AD$ is equally inclined to the coordinate axes,its direction ratios must be equal: $\frac{\lambda - 5}{2} = 1 = \frac{\mu - 8}{2}$.
From $\frac{\lambda - 5}{2} = 1$,we get $\lambda - 5 = 2$,so $\lambda = 7$.
From $\frac{\mu - 8}{2} = 1$,we get $\mu - 8 = 2$,so $\mu = 10$.
Now,calculate $(\lambda^3 + \mu^3 + 5) = 7^3 + 10^3 + 5 = 343 + 1000 + 5 = 1348$.

(D) The given line is $\frac{x - 1}{2} = \frac{y}{2} = \frac{z - 1}{1} = k$.
Any point on this line is given by $(2k + 1, 2k, k + 1)$.
The distance between this point and $(1, 0, 1)$ is $6$ units.
Using the distance formula: $\sqrt{(2k + 1 - 1)^2 + (2k - 0)^2 + (k + 1 - 1)^2} = 6$.
$\sqrt{(2k)^2 + (2k)^2 + k^2} = 6$.
$\sqrt{4k^2 + 4k^2 + k^2} = 6$.
$\sqrt{9k^2} = 6$.
$3|k| = 6$,so $k = \pm 2$.
Since the point lies in the negative $z$-direction relative to the point $(1, 0, 1)$,we look at the $z$-coordinate $z = k + 1$.
For $k = -2$,$z = -2 + 1 = -1$,which is less than $1$.
Substituting $k = -2$ into the coordinates: $x = 2(-2) + 1 = -3$,$y = 2(-2) = -4$,$z = -2 + 1 = -1$.
Thus,the coordinates of $A$ are $(-3, -4, -1)$.

(C) The equation of a line passing through a point $(x_1, y_1, z_1)$ and having direction ratios $(a, b, c)$ is given by $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.
Given that the line passes through the point $(2, -1, 1)$,we have $x_1 = 2, y_1 = -1, z_1 = 1$.
Since the required line is parallel to the line $\frac{x - 3}{2} = \frac{y + 1}{7} = \frac{z - 2}{-3}$,it will have the same direction ratios as the given line.
Thus,the direction ratios are $a = 2, b = 7, c = -3$.
Substituting these values into the standard equation,we get $\frac{x - 2}{2} = \frac{y - (-1)}{7} = \frac{z - 1}{-3}$,which simplifies to $\frac{x - 2}{2} = \frac{y + 1}{7} = \frac{z - 1}{-3}$.

(B) The line $L$ passes through $(1, 2, 3)$,so $\frac{1 - 2}{2} = \frac{2 - 1}{b} = \frac{3 + 1}{c} \Rightarrow -\frac{1}{2} = \frac{1}{b} = \frac{4}{c}$.
Thus,$b = -2$ and $c = -8$.
The line $L$ has direction vector $\vec{v} = (2, -2, -8)$.
Since line $K$ is parallel to $L$,its direction vector must be proportional to $(2, -2, -8)$.
Given $K: \frac{x + 2}{a} = \frac{y - 3}{2} = \frac{z + 4}{d}$,we have $\frac{a}{2} = \frac{2}{-2} = \frac{d}{-8} \Rightarrow \frac{a}{2} = -1 = \frac{d}{-8}$.
Thus,$a = -2$ and $d = 8$.
Line $L$ passes through $P_1 = (2, 1, -1)$ with direction $\vec{v} = (2, -2, -8)$.
Line $K$ passes through $P_2 = (-2, 3, -4)$ with direction $\vec{v} = (2, -2, -8)$.
The distance $d$ between parallel lines is given by $d = \frac{|\vec{P_1P_2} \times \vec{v}|}{|\vec{v}|}$.
$\vec{P_1P_2} = (-2 - 2, 3 - 1, -4 - (-1)) = (-4, 2, -3)$.
$\vec{P_1P_2} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 2 & -3 \\ 2 & -2 & -8 \end{vmatrix} = \hat{i}(-16 - 6) - \hat{j}(32 + 6) + \hat{k}(8 - 4) = (-22, -38, 4)$.
$|\vec{P_1P_2} \times \vec{v}| = \sqrt{(-22)^2 + (-38)^2 + 4^2} = \sqrt{484 + 1444 + 16} = \sqrt{1944}$.
$|\vec{v}| = \sqrt{2^2 + (-2)^2 + (-8)^2} = \sqrt{4 + 4 + 64} = \sqrt{72}$.
Distance $= \frac{\sqrt{1944}}{\sqrt{72}} = \sqrt{\frac{1944}{72}} = \sqrt{27} = 3\sqrt{3} = \sqrt{27} = \frac{\sqrt{243}}{3}$.

(A) Let the given point be $P(2, 1, 3)$ and the line be $L: \vec{r} = (0, 1, -2) + \lambda(1, 1, -1)$.
Any point $Q$ on the line is given by $Q(\lambda, 1+\lambda, -2-\lambda)$.
The vector $\vec{PQ} = (\lambda-2, \lambda, -5-\lambda)$.
Since $PQ$ is perpendicular to the line direction $\vec{v} = (1, 1, -1)$,their dot product is zero:
$(\lambda-2)(1) + (\lambda)(1) + (-5-\lambda)(-1) = 0$
$\lambda - 2 + \lambda + 5 + \lambda = 0$
$3\lambda + 3 = 0 \implies \lambda = -1$.
Substituting $\lambda = -1$ into $Q$,we get the foot of the perpendicular $M(-1, 0, -1)$.
Let the image of $P$ be $P'(x, y, z)$. Since $M$ is the midpoint of $PP'$,we have:
$M = \frac{P + P'}{2} \implies (-1, 0, -1) = \frac{(2, 1, 3) + (x, y, z)}{2}$
$-2 = 2 + x \implies x = -4$
$0 = 1 + y \implies y = -1$
$-2 = 3 + z \implies z = -5$
Thus,the position vector of the image is $-4\hat{i} - \hat{j} - 5\hat{k}$.

(B) The given lines are $L_1: \vec{r} = (3\hat{i} - 15\hat{j} + 9\hat{k}) + \lambda(2\hat{i} - 7\hat{j} + 5\hat{k})$ and $L_2: \vec{r} = (-1\hat{i} + 1\hat{j} + 9\hat{k}) + \mu(2\hat{i} + 1\hat{j} - 3\hat{k})$.
Here,$\vec{a}_1 = 3\hat{i} - 15\hat{j} + 9\hat{k}$,$\vec{a}_2 = -\hat{i} + \hat{j} + 9\hat{k}$,$\vec{b}_1 = 2\hat{i} - 7\hat{j} + 5\hat{k}$,and $\vec{b}_2 = 2\hat{i} + \hat{j} - 3\hat{k}$.
First,find $\vec{a}_2 - \vec{a}_1 = (-1 - 3)\hat{i} + (1 - (-15))\hat{j} + (9 - 9)\hat{k} = -4\hat{i} + 16\hat{j} + 0\hat{k}$.
Next,find the cross product $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = \hat{i}(21 - 5) - \hat{j}(-6 - 10) + \hat{k}(2 + 14) = 16\hat{i} + 16\hat{j} + 16\hat{k}$.
The magnitude $|\vec{b}_1 \times \vec{b}_2| = \sqrt{16^2 + 16^2 + 16^2} = 16\sqrt{3}$.
The shortest distance $SD = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} = \frac{|(-4)(16) + (16)(16) + (0)(16)|}{16\sqrt{3}} = \frac{|-64 + 256|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}$.

(D) Let the points be $A(-1, 3, 2)$,$B(-4, 2, -2)$,and $C(5, 5, \lambda)$.
Since the points are collinear,the vector $\vec{AB}$ must be proportional to the vector $\vec{BC}$.
$\vec{AB} = (-4 - (-1))\hat{i} + (2 - 3)\hat{j} + (-2 - 2)\hat{k} = -3\hat{i} - 1\hat{j} - 4\hat{k}$.
$\vec{BC} = (5 - (-4))\hat{i} + (5 - 2)\hat{j} + (\lambda - (-2))\hat{k} = 9\hat{i} + 3\hat{j} + (\lambda + 2)\hat{k}$.
For collinearity,the ratios of the components must be equal:
$\frac{-3}{9} = \frac{-1}{3} = \frac{-4}{\lambda + 2}$.
From the equality $\frac{-1}{3} = \frac{-4}{\lambda + 2}$,we get:
$-1(\lambda + 2) = -12$.
$\lambda + 2 = 12$.
$\lambda = 10$.

(D) Let the points on the two lines be $P(3\lambda + 6, -\lambda + 7, \lambda + 4)$ and $Q(-3\mu, 2\mu - 9, 4\mu + 2)$.
The vector $\vec{PQ} = (-3\mu - 3\lambda - 6)\hat{i} + (2\mu + \lambda - 16)\hat{j} + (4\mu - \lambda - 2)\hat{k}$.
Since $\vec{PQ}$ is perpendicular to both lines,its dot product with the direction vectors of the lines must be zero.
Direction vectors are $\vec{d_1} = 3\hat{i} - \hat{j} + \hat{k}$ and $\vec{d_2} = -3\hat{i} + 2\hat{j} + 4\hat{k}$.
$\vec{PQ} \cdot \vec{d_1} = 0 \Rightarrow 3(-3\mu - 3\lambda - 6) - 1(2\mu + \lambda - 16) + 1(4\mu - \lambda - 2) = 0 \Rightarrow -7\mu - 11\lambda = 4$.
$\vec{PQ} \cdot \vec{d_2} = 0 \Rightarrow -3(-3\mu - 3\lambda - 6) + 2(2\mu + \lambda - 16) + 4(4\mu - \lambda - 2) = 0 \Rightarrow 29\mu + 7\lambda = 22$.
Solving the system of equations,we get $\lambda = -1$ and $\mu = 1$.
Substituting these values,we get $P(3, 8, 3)$ and $Q(-3, -7, 6)$.
The direction ratios of line $PQ$ are $(-3 - 3, -7 - 8, 6 - 3) = (-6, -15, 3)$,which simplifies to $(2, 5, -1)$.
The equation of the line passing through $P(3, 8, 3)$ with direction ratios $(2, 5, -1)$ is $\frac{x - 3}{2} = \frac{y - 8}{5} = \frac{z - 3}{-1}$.

(A) The given lines are $L_1: \frac{x - 1}{3} = \frac{y - 2}{-1} = \frac{z - \lambda}{2}$ and $L_2: \frac{x + 1}{-2} = \frac{y}{3\lambda} = \frac{z - 7/2}{1/2}$.
For two lines to be coplanar,the determinant of the vector connecting a point on each line and the direction vectors of the lines must be zero:
$\left|\begin{array}{ccc} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right| = 0$
Substituting the points $(1, 2, \lambda)$ and $(-1, 0, 7/2)$ and direction vectors $(3, -1, 2)$ and $(-2, 3\lambda, 1/2)$:
$\left|\begin{array}{ccc} -1 - 1 & 0 - 2 & 7/2 - \lambda \\ 3 & -1 & 2 \\ -2 & 3\lambda & 1/2 \end{array}\right| = 0$
$\left|\begin{array}{ccc} -2 & -2 & (7-2\lambda)/2 \\ 3 & -1 & 2 \\ -2 & 3\lambda & 1/2 \end{array}\right| = 0$
Expanding the determinant:
$-2(-1/2 - 6\lambda) + 2(3/2 + 4) + \frac{7-2\lambda}{2}(9\lambda - 2) = 0$
$1 + 12\lambda + 11 + \frac{63\lambda - 14 - 18\lambda^2 + 4\lambda}{2} = 0$
$12 + 12\lambda + \frac{-18\lambda^2 + 67\lambda - 14}{2} = 0$
$24 + 24\lambda - 18\lambda^2 + 67\lambda - 14 = 0$
$-18\lambda^2 + 91\lambda + 10 = 0 \Rightarrow 18\lambda^2 - 91\lambda - 10 = 0$
Sum of values of $\lambda = -(-91)/18 = 91/18 = 182/36$.

(B) Given equations are:
$l + 3m + 5n = 0$ ....$(1)$
$5lm - 2mn + 6nl = 0$ ....$(2)$
From equation $(1)$,we have $l = -3m - 5n$.
Substituting the value of $l$ in equation $(2)$:
$5(-3m - 5n)m - 2mn + 6n(-3m - 5n) = 0$
$-15m^2 - 25mn - 2mn - 18mn - 30n^2 = 0$
$-15m^2 - 45mn - 30n^2 = 0$
Dividing by $-15$,we get:
$m^2 + 3mn + 2n^2 = 0$
$(m + n)(m + 2n) = 0$
So,$m = -n$ or $m = -2n$.
Case $1$: If $m = -n$,then $l = -3(-n) - 5n = 3n - 5n = -2n$. The direction ratios are $(-2n, -n, n)$,which simplifies to $(-2, -1, 1)$.
Case $2$: If $m = -2n$,then $l = -3(-2n) - 5n = 6n - 5n = n$. The direction ratios are $(n, -2n, n)$,which simplifies to $(1, -2, 1)$.
Let the direction ratios be $\vec{a} = (-2, -1, 1)$ and $\vec{b} = (1, -2, 1)$.
The angle $\theta$ between the lines is given by:
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(-2)(1) + (-1)(-2) + (1)(1)|}{\sqrt{(-2)^2 + (-1)^2 + 1^2} \sqrt{1^2 + (-2)^2 + 1^2}}$
$\cos \theta = \frac{|-2 + 2 + 1|}{\sqrt{6} \sqrt{6}} = \frac{1}{6}$
Therefore,$\theta = \cos^{-1}\left(\frac{1}{6}\right)$.

(D) The direction ratios of the first line $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ are $\vec{v_1} = (2, 2, 1)$.
The second line is $\frac{-(x - 5)}{-2} = \frac{7(y - 2)}{p} = \frac{z - 3}{4}$,which simplifies to $\frac{x - 5}{2} = \frac{y - 2}{p/7} = \frac{z - 3}{4}$.
The direction ratios of the second line are $\vec{v_2} = (2, p/7, 4)$.
The angle $\theta$ between the lines is given by $\cos \theta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.
Given $\cos \theta = \frac{2}{3}$,we have $\frac{2}{3} = \frac{|2(2) + 2(p/7) + 1(4)|}{\sqrt{2^2 + 2^2 + 1^2} \sqrt{2^2 + (p/7)^2 + 4^2}}$.
$\frac{2}{3} = \frac{|8 + 2p/7|}{3 \sqrt{4 + p^2/49 + 16}} = \frac{|8 + 2p/7|}{3 \sqrt{20 + p^2/49}}$.
Canceling $3$ from the denominator,we get $\frac{2}{1} = \frac{|8 + 2p/7|}{\sqrt{20 + p^2/49}}$.
Squaring both sides: $4 = \frac{(8 + 2p/7)^2}{20 + p^2/49} = \frac{64 + 32p/7 + 4p^2/49}{20 + p^2/49}$.
$80 + 4p^2/49 = 64 + 32p/7 + 4p^2/49$.
$80 = 64 + 32p/7 \Rightarrow 16 = 32p/7 \Rightarrow p = \frac{16 \times 7}{32} = \frac{7}{2}$.

(B) The shortest distance $d$ between two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ is given by the formula:
$d = \frac{|(x_2-x_1, y_2-y_1, z_2-z_1) \cdot (a_1, b_1, c_1) \times (a_2, b_2, c_2)|}{|(a_1, b_1, c_1) \times (a_2, b_2, c_2)|}$
For the given lines,we have $(x_1, y_1, z_1) = (0, 0, 0)$ and $(a_1, b_1, c_1) = (2, 2, 1)$.
For the second line,$(x_2, y_2, z_2) = (-2, 4, 5)$ and $(a_2, b_2, c_2) = (-1, 8, 4)$.
The vector product $(a_1, b_1, c_1) \times (a_2, b_2, c_2) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ -1 & 8 & 4 \end{vmatrix} = \hat{i}(8-8) - \hat{j}(8+1) + \hat{k}(16+2) = (0, -9, 18)$.
The magnitude is $\sqrt{0^2 + (-9)^2 + 18^2} = \sqrt{81 + 324} = \sqrt{405} = 9\sqrt{5}$.
The scalar triple product is $\begin{vmatrix} -2 & 4 & 5 \\ 2 & 2 & 1 \\ -1 & 8 & 4 \end{vmatrix} = -2(8-8) - 4(8+1) + 5(16+2) = 0 - 36 + 90 = 54$.
Thus,$d = \frac{|54|}{9\sqrt{5}} = \frac{6}{\sqrt{5}} = \frac{6 \times 2.236}{5} \approx 2.68$.
Since $2.68 \in (2, 3]$,the correct option is $B$.

(B) Let the two lines be $L_1: \frac{x}{1} = \frac{y}{-1} = \frac{z}{1} = r_1$ and $L_2: \frac{x - 1}{0} = \frac{y + 1}{-2} = \frac{z}{1} = r_2$.
Any point on $L_1$ is $P(r_1, -r_1, r_1)$ and any point on $L_2$ is $Q(1, -2r_2 - 1, r_2)$.
The direction ratios of the line $PQ$ are $(1 - r_1, -2r_2 - 1 + r_1, r_2 - r_1)$.
Since $PQ$ is the shortest distance line,it is perpendicular to both $L_1$ (direction $\vec{v_1} = \langle 1, -1, 1 \rangle$) and $L_2$ (direction $\vec{v_2} = \langle 0, -2, 1 \rangle$).
$1$) $(1 - r_1)(1) + (-2r_2 - 1 + r_1)(-1) + (r_2 - r_1)(1) = 0 \Rightarrow 1 - r_1 + 2r_2 + 1 - r_1 + r_2 - r_1 = 0 \Rightarrow -3r_1 + 3r_2 + 2 = 0$.
$2$) $(1 - r_1)(0) + (-2r_2 - 1 + r_1)(-2) + (r_2 - r_1)(1) = 0 \Rightarrow 4r_2 + 2 - 2r_1 + r_2 - r_1 = 0 \Rightarrow -3r_1 + 5r_2 + 2 = 0$.
Subtracting the equations: $2r_2 = 0 \Rightarrow r_2 = 0$.
Then $-3r_1 + 2 = 0 \Rightarrow r_1 = 2/3$.
The points are $P(2/3, -2/3, 2/3)$ and $Q(1, -1, 0)$.
The direction ratios of $PQ$ are $(1 - 2/3, -1 + 2/3, 0 - 2/3) = (1/3, -1/3, -2/3)$,which is proportional to $(1, -1, -2)$.
The line passes through $Q(1, -1, 0)$,so the equation is $\frac{x - 1}{1} = \frac{y + 1}{-1} = \frac{z}{-2}$.

(A) Two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ are coplanar if $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$.
Here,$(x_1, y_1, z_1) = (-1, 1, -1)$ and $(x_2, y_2, z_2) = (-2, k, 0)$.
The direction ratios are $(a_1, b_1, c_1) = (2, 1, 3)$ and $(a_2, b_2, c_2) = (2, 3, 4)$.
Substituting these values into the determinant:
$\begin{vmatrix} -2 - (-1) & k - 1 & 0 - (-1) \\ 2 & 1 & 3 \\ 2 & 3 & 4 \end{vmatrix} = 0$
$\Rightarrow \begin{vmatrix} -1 & k - 1 & 1 \\ 2 & 1 & 3 \\ 2 & 3 & 4 \end{vmatrix} = 0$
Expanding along the first row:
$-1(4 - 9) - (k - 1)(8 - 6) + 1(6 - 2) = 0$
$-1(-5) - (k - 1)(2) + 4 = 0$
$5 - 2k + 2 + 4 = 0$
$11 - 2k = 0$
$2k = 11$
$k = \frac{11}{2}$.

(D) For line $L_1$,we have $x = \sqrt{\lambda} y + (\sqrt{\lambda} - 1)$ and $z = (\sqrt{\lambda} - 1)y + \sqrt{\lambda}$.
Rewriting these in symmetric form,we get $\frac{x - (\sqrt{\lambda} - 1)}{\sqrt{\lambda}} = y = \frac{z - \sqrt{\lambda}}{\sqrt{\lambda} - 1}$.
The direction vector of $L_1$ is $\vec{v_1} = (\sqrt{\lambda}, 1, \sqrt{\lambda} - 1)$.
Similarly,for line $L_2$,we have $x = \sqrt{\mu} y + (1 - \sqrt{\mu})$ and $z = (1 - \sqrt{\mu})y + \sqrt{\mu}$.
Rewriting these in symmetric form,we get $\frac{x - (1 - \sqrt{\mu})}{\sqrt{\mu}} = y = \frac{z - \sqrt{\mu}}{1 - \sqrt{\mu}}$.
The direction vector of $L_2$ is $\vec{v_2} = (\sqrt{\mu}, 1, 1 - \sqrt{\mu})$.
Since $L_1 \perp L_2$,the dot product of their direction vectors must be zero: $\vec{v_1} \cdot \vec{v_2} = 0$.
$(\sqrt{\lambda})(\sqrt{\mu}) + (1)(1) + (\sqrt{\lambda} - 1)(1 - \sqrt{\mu}) = 0$.
$\sqrt{\lambda\mu} + 1 + (\sqrt{\lambda} - \sqrt{\lambda\mu} - 1 + \sqrt{\mu}) = 0$.
$\sqrt{\lambda} + \sqrt{\mu} = 0$.
Since $\lambda, \mu \geq 0$,this implies $\sqrt{\lambda} = 0$ and $\sqrt{\mu} = 0$,which means $\lambda = 0$ and $\mu = 0$. Thus,$\lambda = \mu$.

(D) Let the given line be $\frac{x - 1}{2} = \frac{y + 1}{-3} = \frac{z + 10}{8} = k$.
Any point $L$ on the line is given by $L = (2k + 1, -3k - 1, 8k - 10)$.
Let $P = (1, 0, 0)$. The direction ratios of the line $PL$ are $(2k + 1 - 1, -3k - 1 - 0, 8k - 10 - 0) = (2k, -3k - 1, 8k - 10)$.
Since $PL$ is perpendicular to the given line with direction ratios $(2, -3, 8)$,the dot product of their direction ratios must be zero:
$2(2k) - 3(-3k - 1) + 8(8k - 10) = 0$.
$4k + 9k + 3 + 64k - 80 = 0$.
$77k - 77 = 0$,which gives $k = 1$.
Substituting $k = 1$ into the coordinates of $L$:
$L = (2(1) + 1, -3(1) - 1, 8(1) - 10) = (3, -4, -2)$.

(B) The given lines are $L_1: \frac{x}{2} = \frac{y}{-1} = \frac{z}{2}$ and $L_2: \frac{x-1}{4} = \frac{y-1}{-2} = \frac{z-1}{4}$.
Both lines have the same direction vector $\vec{b} = 2\hat{i} - \hat{j} + 2\hat{k}$,so they are parallel.
The formula for the shortest distance between two parallel lines $\vec{r} = \vec{a}_1 + \lambda\vec{b}$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}$ is $d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}$.
Here,$\vec{a}_1 = 0\hat{i} + 0\hat{j} + 0\hat{k}$ and $\vec{a}_2 = 1\hat{i} + 1\hat{j} + 1\hat{k}$.
Thus,$\vec{a}_2 - \vec{a}_1 = \hat{i} + \hat{j} + \hat{k}$.
$(\vec{a}_2 - \vec{a}_1) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 2 \end{vmatrix} = \hat{i}(2 - (-1)) - \hat{j}(2 - 2) + \hat{k}(-1 - 2) = 3\hat{i} + 0\hat{j} - 3\hat{k}$.
The magnitude is $|3\hat{i} - 3\hat{k}| = \sqrt{3^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}$.
The magnitude of $\vec{b}$ is $|2\hat{i} - \hat{j} + 2\hat{k}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3$.
Therefore,$d = \frac{3\sqrt{2}}{3} = \sqrt{2}$.
Since the calculated distance is $\sqrt{2}$,Statement $1$ is true.
Statement $2$ is the standard definition of the shortest distance between parallel lines,which is also true.
Since Statement $2$ provides the method used to calculate the distance in Statement $1$,it is the correct explanation.

(C) Let the given point be $P(-1, 2, 6)$ and the line $L$ pass through $A(2, 3, -4)$ with direction vector $\vec{v} = 6\hat{i} + 3\hat{j} - 4\hat{k}$.
The vector $\vec{AP} = (-1-2)\hat{i} + (2-3)\hat{j} + (6-(-4))\hat{k} = -3\hat{i} - \hat{j} + 10\hat{k}$.
The distance $d$ of point $P$ from the line is given by $d = \frac{|\vec{AP} \times \vec{v}|}{|\vec{v}|}$.
First,calculate the cross product $\vec{AP} \times \vec{v}$:
$\vec{AP} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & -1 & 10 \\ 6 & 3 & -4 \end{vmatrix} = \hat{i}(4 - 30) - \hat{j}(12 - 60) + \hat{k}(-9 + 6) = -26\hat{i} + 48\hat{j} - 3\hat{k}$.
The magnitude $|\vec{AP} \times \vec{v}| = \sqrt{(-26)^2 + 48^2 + (-3)^2} = \sqrt{676 + 2304 + 9} = \sqrt{2989}$.
The magnitude $|\vec{v}| = \sqrt{6^2 + 3^2 + (-4)^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$.
Thus,$d = \sqrt{\frac{2989}{61}} = \sqrt{49} = 7$.

(B) The first line is given by $x = ay + b$ and $z = cy + d$.
This can be written in symmetric form as $\frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c}$.
The direction ratios of this line are $(a, 1, c)$.
The second line is given by $x = a'z + b'$ and $y = c'z + d'$.
This can be written in symmetric form as $\frac{x - b'}{a'} = \frac{y - d'}{c'} = \frac{z}{1}$.
The direction ratios of this line are $(a', c', 1)$.
Since the two lines are perpendicular,the dot product of their direction ratios must be zero:
$(a)(a') + (1)(c') + (c)(1) = 0$.
Therefore,$aa' + c' + c = 0$.

(A) Let the points on the two lines be $P_1 = (\lambda + 3, 3\lambda - 1, -\lambda + 6)$ and $P_2 = (7\alpha - 5, -6\alpha + 2, 4\alpha + 3)$.
For the lines to intersect at point $R$,we equate the coordinates:
$\lambda + 3 = 7\alpha - 5 \Rightarrow \lambda - 7\alpha = -8$ (Equation $1$)
$3\lambda - 1 = -6\alpha + 2 \Rightarrow 3\lambda + 6\alpha = 3 \Rightarrow \lambda + 2\alpha = 1$ (Equation $2$)
Subtracting Equation $1$ from Equation $2$ gives $9\alpha = 9$,so $\alpha = 1$.
Substituting $\alpha = 1$ into Equation $2$,we get $\lambda + 2(1) = 1$,so $\lambda = -1$.
Substituting $\lambda = -1$ into the first line equation,we get $R = (-1 + 3, 3(-1) - 1, -(-1) + 6) = (2, -4, 7)$.
The reflection of a point $(x, y, z)$ in the $xy$-plane is $(x, y, -z)$.
Therefore,the reflection of $R(2, -4, 7)$ in the $xy$-plane is $(2, -4, -7)$.

(D) Let the given line be $\frac{x + 3}{10} = \frac{y - 2}{-7} = \frac{z}{1} = \lambda$.
Any point $M$ on the line is given by $(10\lambda - 3, -7\lambda + 2, \lambda)$.
The vector $\vec{PM} = (10\lambda - 3 - 2, -7\lambda + 2 - (-1), \lambda - 4) = (10\lambda - 5, -7\lambda + 3, \lambda - 4)$.
Since $\vec{PM}$ is perpendicular to the line with direction ratios $(10, -7, 1)$,we have:
$10(10\lambda - 5) - 7(-7\lambda + 3) + 1(\lambda - 4) = 0$
$100\lambda - 50 + 49\lambda - 21 + \lambda - 4 = 0$
$150\lambda - 75 = 0 \Rightarrow \lambda = \frac{1}{2}$.
The coordinates of $M$ are $(10(\frac{1}{2}) - 3, -7(\frac{1}{2}) + 2, \frac{1}{2}) = (2, -1.5, 0.5)$.
The length of the perpendicular $PM$ is $\sqrt{(2-2)^2 + (-1.5 - (-1))^2 + (0.5 - 4)^2}$
$= \sqrt{0^2 + (-0.5)^2 + (-3.5)^2} = \sqrt{0.25 + 12.25} = \sqrt{12.5} = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}}$.
Since $\sqrt{2} \approx 1.414$,$\frac{5}{1.414} \approx 3.535$.
This value is greater than $3$ but less than $4$.

(C) The equation of the line passing through $P(2, -3, 4)$ and $Q(8, 0, 10)$ is given by $\frac{x-2}{8-2} = \frac{y-(-3)}{0-(-3)} = \frac{z-4}{10-4}$.
This simplifies to $\frac{x-2}{6} = \frac{y+3}{3} = \frac{z-4}{6}$.
Since point $R(4, y, z)$ lies on this line,we substitute $x=4$ into the equation:
$\frac{4-2}{6} = \frac{y+3}{3} = \frac{z-4}{6}$.
$\frac{2}{6} = \frac{1}{3} = \frac{y+3}{3} = \frac{z-4}{6}$.
From $\frac{1}{3} = \frac{y+3}{3}$,we get $y+3 = 1$,so $y = -2$.
From $\frac{1}{3} = \frac{z-4}{6}$,we get $z-4 = 2$,so $z = 6$.
Thus,the coordinates of $R$ are $(4, -2, 6)$.
The distance of $R(4, -2, 6)$ from the origin $(0, 0, 0)$ is $\sqrt{4^2 + (-2)^2 + 6^2} = \sqrt{16 + 4 + 36} = \sqrt{56} = 2\sqrt{14}$.

(B) Let the line be $L: \frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4} = \lambda$. Any point on this line is $P(3\lambda - 2, 1, 4\lambda)$.
Let $D$ be the foot of the perpendicular from $A(1, -1, 2)$ to the line $L$. Since $D$ lies on $L$,$D = (3\lambda - 2, 1, 4\lambda)$.
The direction vector of the line $L$ is $\vec{v} = 3\hat{i} + 0\hat{j} + 4\hat{k}$.
The vector $\vec{AD} = (3\lambda - 2 - 1)\hat{i} + (1 - (-1))\hat{j} + (4\lambda - 2)\hat{k} = (3\lambda - 3)\hat{i} + 2\hat{j} + (4\lambda - 2)\hat{k}$.
Since $\vec{AD} \perp \vec{v}$,their dot product is zero:
$3(3\lambda - 3) + 0(2) + 4(4\lambda - 2) = 0$
$9\lambda - 9 + 16\lambda - 8 = 0$
$25\lambda = 17 \implies \lambda = \frac{17}{25}$.
Substituting $\lambda$ back into $\vec{AD}$:
$\vec{AD} = (3(\frac{17}{25}) - 3)\hat{i} + 2\hat{j} + (4(\frac{17}{25}) - 2)\hat{k} = (\frac{51-75}{25})\hat{i} + 2\hat{j} + (\frac{68-50}{25})\hat{k} = -\frac{24}{25}\hat{i} + 2\hat{j} + \frac{18}{25}\hat{k}$.
The length of the altitude $AD = |\vec{AD}| = \sqrt{(-\frac{24}{25})^2 + 2^2 + (\frac{18}{25})^2} = \sqrt{\frac{576}{625} + 4 + \frac{324}{625}} = \sqrt{\frac{900}{625} + 4} = \sqrt{\frac{36}{25} + 4} = \sqrt{\frac{36+100}{25}} = \sqrt{\frac{136}{25}} = \frac{\sqrt{4 \times 34}}{5} = \frac{2\sqrt{34}}{5}$.
The area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times 5 \times \frac{2\sqrt{34}}{5} = \sqrt{34}$.

(B) Let the line be $\frac{x}{1} = \frac{y - 1}{0} = \frac{z + 1}{-1} = \lambda$.
Any point $C$ on the line is given by $(\lambda, 1, -\lambda - 1)$.
The given point is $P(\beta, 0, \beta)$.
The direction vector of the line is $\vec{v} = (1, 0, -1)$.
The vector $\vec{PC} = (\lambda - \beta, 1 - 0, -\lambda - 1 - \beta) = (\lambda - \beta, 1, -\lambda - \beta - 1)$.
Since $PC$ is perpendicular to the line,$\vec{PC} \cdot \vec{v} = 0$:
$(\lambda - \beta)(1) + (1)(0) + (-\lambda - \beta - 1)(-1) = 0$
$\lambda - \beta + \lambda + \beta + 1 = 0$
$2\lambda + 1 = 0 \implies \lambda = -\frac{1}{2}$.
Thus,the point $C$ is $(-\frac{1}{2}, 1, -\frac{1}{2})$.
The length of the perpendicular $PC$ is $\sqrt{(\beta - (-\frac{1}{2}))^2 + (0 - 1)^2 + (\beta - (-\frac{1}{2}))^2} = \sqrt{\frac{3}{2}}$.
Squaring both sides: $(\beta + \frac{1}{2})^2 + 1 + (\beta + \frac{1}{2})^2 = \frac{3}{2}$.
$2(\beta + \frac{1}{2})^2 = \frac{3}{2} - 1 = \frac{1}{2}$.
$(\beta + \frac{1}{2})^2 = \frac{1}{4}$.
$\beta + \frac{1}{2} = \pm \frac{1}{2}$.
Case $1$: $\beta + \frac{1}{2} = \frac{1}{2} \implies \beta = 0$ (Rejected as $\beta \neq 0$).
Case $2$: $\beta + \frac{1}{2} = -\frac{1}{2} \implies \beta = -1$.
Therefore,$\beta = -1$.

(A) Let the given point be $P(-1, 2, 6)$ and the point on the line be $A(2, 3, -4)$. The vector $\vec{AP}$ is given by:
$\vec{AP} = (-1 - 2)\hat{i} + (2 - 3)\hat{j} + (6 - (-4))\hat{k} = -3\hat{i} - \hat{j} + 10\hat{k}$.
The line is parallel to the vector $\vec{b} = 6\hat{i} + 3\hat{j} - 4\hat{k}$.
The distance $d$ of point $P$ from the line is given by the formula:
$d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}$.
First,calculate the cross product $\vec{AP} \times \vec{b}$:
$\vec{AP} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & -1 & 10 \\ 6 & 3 & -4 \end{vmatrix} = \hat{i}(4 - 30) - \hat{j}(12 - 60) + \hat{k}(-9 - (-6)) = -26\hat{i} + 48\hat{j} - 3\hat{k}$.
Now,calculate the magnitude $|\vec{AP} \times \vec{b}| = \sqrt{(-26)^2 + 48^2 + (-3)^2} = \sqrt{676 + 2304 + 9} = \sqrt{2989}$.
Calculate the magnitude $|\vec{b}| = \sqrt{6^2 + 3^2 + (-4)^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$.
Thus,$d = \sqrt{\frac{2989}{61}} = \sqrt{49} = 7$.

(B) Let $A = (1, 0, 3)$,$B = (\alpha, 7, 1)$,and $P = \left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$.
Since $P$ is the foot of the perpendicular from $A$ to the line passing through $B$,the vector $\vec{AP}$ must be perpendicular to the vector $\vec{BP}$.
First,find the direction ratios of $\vec{AP}$:
$\vec{AP} = \left(\frac{5}{3} - 1, \frac{7}{3} - 0, \frac{17}{3} - 3\right) = \left(\frac{2}{3}, \frac{7}{3}, \frac{8}{3}\right)$.
Next,find the direction ratios of $\vec{BP}$:
$\vec{BP} = \left(\frac{5}{3} - \alpha, \frac{7}{3} - 7, \frac{17}{3} - 1\right) = \left(\frac{5}{3} - \alpha, -\frac{14}{3}, \frac{14}{3}\right)$.
Since $\vec{AP} \perp \vec{BP}$,their dot product must be zero:
$\vec{AP} \cdot \vec{BP} = 0$
$\left(\frac{2}{3}\right)\left(\frac{5}{3} - \alpha\right) + \left(\frac{7}{3}\right)\left(-\frac{14}{3}\right) + \left(\frac{8}{3}\right)\left(\frac{14}{3}\right) = 0$
$\frac{2}{3} \left(\frac{5}{3} - \alpha\right) - \frac{98}{9} + \frac{112}{9} = 0$
$\frac{2}{3} \left(\frac{5}{3} - \alpha\right) + \frac{14}{9} = 0$
Multiply by $9$:
$6 \left(\frac{5}{3} - \alpha\right) + 14 = 0$
$10 - 6\alpha + 14 = 0$
$24 = 6\alpha$
$\alpha = 4$.

(B) The lines are given by $\vec{r_1} = (3, 8, 3) + \lambda(3, -1, 1)$ and $\vec{r_2} = (-3, -7, 6) + \mu(-3, 2, 4)$.
Let $\vec{a_1} = (3, 8, 3)$,$\vec{a_2} = (-3, -7, 6)$,$\vec{b_1} = (3, -1, 1)$,and $\vec{b_2} = (-3, 2, 4)$.
The shortest distance $d$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{ |\vec{b_1} \times \vec{b_2}| }$.
First,$\vec{a_2} - \vec{a_1} = (-3-3, -7-8, 6-3) = (-6, -15, 3)$.
Next,$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} = \hat{i}(-4-2) - \hat{j}(12 - (-3)) + \hat{k}(6-3) = -6\hat{i} - 15\hat{j} + 3\hat{k}$.
The magnitude $|\vec{b_1} \times \vec{b_2}| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}$.
The dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-6)(-6) + (-15)(-15) + (3)(3) = 36 + 225 + 9 = 270$.
Thus,$d = \frac{|270|}{\sqrt{270}} = \sqrt{270} = 3\sqrt{30}$.

The given points are $A(1, 2, 7)$,$B(2, 6, 3)$,and $C(3, 10, -1)$.
First,we find the vectors $\overrightarrow{AB}$,$\overrightarrow{BC}$,and $\overrightarrow{AC}$:
$\overrightarrow{AB} = (2-1)\hat{i} + (6-2)\hat{j} + (3-7)\hat{k} = \hat{i} + 4\hat{j} - 4\hat{k}$
$\overrightarrow{BC} = (3-2)\hat{i} + (10-6)\hat{j} + (-1-3)\hat{k} = \hat{i} + 4\hat{j} - 4\hat{k}$
$\overrightarrow{AC} = (3-1)\hat{i} + (10-2)\hat{j} + (-1-7)\hat{k} = 2\hat{i} + 8\hat{j} - 8\hat{k}$
Now,calculate the magnitudes of these vectors:
$|\overrightarrow{AB}| = \sqrt{1^2 + 4^2 + (-4)^2} = \sqrt{1 + 16 + 16} = \sqrt{33}$
$|\overrightarrow{BC}| = \sqrt{1^2 + 4^2 + (-4)^2} = \sqrt{1 + 16 + 16} = \sqrt{33}$
$|\overrightarrow{AC}| = \sqrt{2^2 + 8^2 + (-8)^2} = \sqrt{4 + 64 + 64} = \sqrt{132} = 2\sqrt{33}$
Since $|\overrightarrow{AC}| = |\overrightarrow{AB}| + |\overrightarrow{BC}|$ (i.e.,$2\sqrt{33} = \sqrt{33} + \sqrt{33}$),the points $A$,$B$,and $C$ are collinear.

(N/A) The direction ratios of the line segment joining $A(2, 3, -4)$ and $B(1, -2, 3)$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.
$a_1 = 1 - 2 = -1$
$b_1 = -2 - 3 = -5$
$c_1 = 3 - (-4) = 7$
So,the direction ratios of $AB$ are $(-1, -5, 7)$.
The direction ratios of the line segment joining $B(1, -2, 3)$ and $C(3, 8, -11)$ are:
$a_2 = 3 - 1 = 2$
$b_2 = 8 - (-2) = 10$
$c_2 = -11 - 3 = -14$
So,the direction ratios of $BC$ are $(2, 10, -14)$.
We observe that the direction ratios of $BC$ are $-2$ times the direction ratios of $AB$:
$(2, 10, -14) = -2 \times (-1, -5, 7)$.
Since the direction ratios are proportional,the lines $AB$ and $BC$ are parallel.
Because the point $B$ is common to both segments $AB$ and $BC$,the points $A, B,$ and $C$ must lie on the same straight line.
Therefore,the points $A, B,$ and $C$ are collinear.

(N/A) Let the given points be $A(2,3,4), B(-1,-2,1),$ and $C(5,8,7).$
The direction ratios of the line segment $AB$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1).$
Direction ratios of $AB = (-1 - 2, -2 - 3, 1 - 4) = (-3, -5, -3).$
Direction ratios of $BC = (5 - (-1), 8 - (-2), 7 - 1) = (6, 10, 6).$
We observe that the direction ratios of $BC$ are $-2$ times the direction ratios of $AB$:
$(6, 10, 6) = -2 \times (-3, -5, -3).$
Since the direction ratios are proportional,the lines $AB$ and $BC$ are parallel.
Because point $B$ is common to both $AB$ and $BC,$ the points $A, B,$ and $C$ must lie on the same line.
Therefore,the points $(2,3,4), (-1,-2,1),$ and $(5,8,7)$ are collinear.

(A) The position vector of the given point is $\vec{a} = 5 \hat{i} + 2 \hat{j} - 4 \hat{k}$.
The direction vector of the line is $\vec{b} = 3 \hat{i} + 2 \hat{j} - 8 \hat{k}$.
The vector equation of a line passing through point $\vec{a}$ and parallel to vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda \vec{b}$.
Substituting the values,the vector equation is $\vec{r} = (5 \hat{i} + 2 \hat{j} - 4 \hat{k}) + \lambda (3 \hat{i} + 2 \hat{j} - 8 \hat{k})$.
For the Cartesian equation,let $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$.
Then $x \hat{i} + y \hat{j} + z \hat{k} = (5 + 3 \lambda) \hat{i} + (2 + 2 \lambda) \hat{j} + (-4 - 8 \lambda) \hat{k}$.
Equating the components,we get $x = 5 + 3 \lambda$,$y = 2 + 2 \lambda$,and $z = -4 - 8 \lambda$.
Solving for $\lambda$,we get $\lambda = \frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8}$.
Thus,the Cartesian equation is $\frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8}$.

(A) Let $\vec{a}$ and $\vec{b}$ be the position vectors of the points $A(-1, 0, 2)$ and $B(3, 4, 6)$.
Then,$\vec{a} = -\hat{i} + 2\hat{k}$ and $\vec{b} = 3\hat{i} + 4\hat{j} + 6\hat{k}$.
The direction vector of the line is given by $\vec{b} - \vec{a} = (3 - (-1))\hat{i} + (4 - 0)\hat{j} + (6 - 2)\hat{k} = 4\hat{i} + 4\hat{j} + 4\hat{k}$.
The vector equation of a line passing through points with position vectors $\vec{a}$ and $\vec{b}$ is $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$.
Substituting the values,we get $\vec{r} = -\hat{i} + 2\hat{k} + \lambda(4\hat{i} + 4\hat{j} + 4\hat{k})$.

(A) Comparing the given Cartesian equation $\frac{x+3}{2}=\frac{y-5}{4}=\frac{z+6}{2}$ with the standard form $\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$,we get:
$x_{1}=-3, y_{1}=5, z_{1}=-6$ and $a=2, b=4, c=2$.
Thus,the line passes through the point $A(-3, 5, -6)$ and is parallel to the vector $\vec{b} = 2\hat{i} + 4\hat{j} + 2\hat{k}$.
The position vector of the point is $\vec{a} = -3\hat{i} + 5\hat{j} - 6\hat{k}$.
The vector equation of a line passing through point $\vec{a}$ and parallel to vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda\vec{b}$.
Substituting the values,we get $\vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \lambda(2\hat{i} + 4\hat{j} + 2\hat{k})$.