Line Questions in English

(D) Let the point $P$ be $(5, 1, 3)$.
The line passes through point $Q(3, 7, 1)$ and is parallel to the vector $\vec{v} = \hat{j} + \hat{k}$.
The vector $\vec{PQ} = (3-5)\hat{i} + (7-1)\hat{j} + (1-3)\hat{k} = -2\hat{i} + 6\hat{j} - 2\hat{k}$.
The distance $d$ from a point to a line is given by $d = \frac{|\vec{PQ} \times \vec{v}|}{|\vec{v}|}$.
First,calculate the cross product $\vec{PQ} \times \vec{v}$:
$\vec{PQ} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 6 & -2 \\ 0 & 1 & 1 \end{vmatrix} = \hat{i}(6 - (-2)) - \hat{j}(-2 - 0) + \hat{k}(-2 - 0) = 8\hat{i} + 2\hat{j} - 2\hat{k}$.
The magnitude $|\vec{PQ} \times \vec{v}| = \sqrt{8^2 + 2^2 + (-2)^2} = \sqrt{64 + 4 + 4} = \sqrt{72} = 6\sqrt{2}$.
The magnitude $|\vec{v}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$.
Therefore,$d = \frac{6\sqrt{2}}{\sqrt{2}} = 6$.

(D) Given that line $L_1$ passes through $\vec{a}_1 = 2 \hat{i}-\hat{k}$ and is parallel to $\vec{b}_1 = 3 \hat{i}-\hat{j}+2 \hat{k}$.
Line $L_2$ passes through $\vec{a}_2 = 2 \hat{i}+\hat{j}-3 \hat{k}$ and is parallel to $\vec{b}_2 = \hat{i}-2 \hat{j}+\hat{k}$.
The shortest distance $d$ between two skew lines is given by the formula: $d = \frac{|(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}_1 \times \vec{b}_2|}$.
First,calculate the cross product $\vec{b}_1 \times \vec{b}_2$:
$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 2 \\ 1 & -2 & 1 \end{vmatrix} = \hat{i}(-1 + 4) - \hat{j}(3 - 2) + \hat{k}(-6 + 1) = 3 \hat{i} - \hat{j} - 5 \hat{k}$.
The magnitude is $|\vec{b}_1 \times \vec{b}_2| = \sqrt{3^2 + (-1)^2 + (-5)^2} = \sqrt{9 + 1 + 25} = \sqrt{35}$.
Next,calculate $\vec{a}_2 - \vec{a}_1 = (2 \hat{i} + \hat{j} - 3 \hat{k}) - (2 \hat{i} - \hat{k}) = \hat{j} - 2 \hat{k}$.
Now,calculate the dot product: $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = (3 \hat{i} - \hat{j} - 5 \hat{k}) \cdot (0 \hat{i} + 1 \hat{j} - 2 \hat{k}) = (3)(0) + (-1)(1) + (-5)(-2) = 0 - 1 + 10 = 9$.
Therefore,the shortest distance is $d = \frac{|9|}{\sqrt{35}} = \frac{9}{\sqrt{35}}$.

(D) The direction ratios of the line segment $\overline{AB}$ are given by $(4-2, 5-3, 7-4) = (2, 2, 3)$.
The direction ratios of the line segment $\overline{CD}$ are given by $(4-2, -4-(-6), k-3) = (2, 2, k-3)$.
Since the line $\overline{AB}$ is parallel to $\overline{CD}$,their direction ratios must be proportional.
Therefore,$\frac{2}{2} = \frac{2}{2} = \frac{3}{k-3}$.
This implies $1 = \frac{3}{k-3}$.
Solving for $k$,we get $k-3 = 3$,which means $k = 6$.

(B) The direction ratios of the line joining the points $(k, 2, 3)$ and $(1, 1, 2)$ are $(1-k, 1-2, 2-3)$,which simplifies to $(1-k, -1, -1)$.
The direction ratios of the line joining the points $(5, 4, -1)$ and $(3, 2, -3)$ are $(3-5, 2-4, -3-(-1))$,which simplifies to $(-2, -2, -2)$.
Since the two lines are parallel,their direction ratios must be proportional:
$\frac{1-k}{-2} = \frac{-1}{-2} = \frac{-1}{-2}$
From the equality $\frac{1-k}{-2} = \frac{1}{2}$,we get:
$1-k = -1$
$k = 2$
Therefore,the value of $k$ is $2$.

(A) Let the direction cosines of the lines be $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$.
Given equations are $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$.
From the first equation,$l = -\frac{bm + cn}{a}$. Substituting this into the second equation:
$fmn + gn(-\frac{bm + cn}{a}) + hm(-\frac{bm + cn}{a}) = 0$
$afmn - bgnm - cgn^2 - bhm^2 - chmn = 0$
$bhm^2 + (ch + bg - af)mn + cgn^2 = 0$
Dividing by $n^2$,we get $bh(\frac{m}{n})^2 + (ch + bg - af)(\frac{m}{n}) + cg = 0$.
The product of the roots is $\frac{m_1 m_2}{n_1 n_2} = \frac{cg}{bh}$.
Similarly,by eliminating $m$,we get $ah(\frac{l}{n})^2 + (ch + af - bg)(\frac{l}{n}) + cf = 0$,so $\frac{l_1 l_2}{n_1 n_2} = \frac{cf}{ah}$.
Since the lines are perpendicular,$l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.
Dividing by $n_1 n_2$,we get $\frac{l_1 l_2}{n_1 n_2} + \frac{m_1 m_2}{n_1 n_2} + 1 = 0$.
Substituting the products: $\frac{cf}{ah} + \frac{cg}{bh} + 1 = 0$.
Dividing by $c$,we get $\frac{f}{a} + \frac{g}{b} + \frac{h}{c} = 0$.

(C) The given Cartesian equation is $2x - 3 = 3y + 1 = 5 - 6z$.
To write this in standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$,we divide by the coefficients of $x, y, z$:
$2(x - \frac{3}{2}) = 3(y + \frac{1}{3}) = -6(z - \frac{5}{6})$.
Dividing by $6$,we get $\frac{x - 3/2}{3} = \frac{y + 1/3}{2} = \frac{z - 5/6}{-1}$.
The direction vector of this line is $\vec{v} = 3 \hat{i} + 2 \hat{j} - \hat{k}$.
The vector equation of a line passing through point $\vec{a} = 7 \hat{i} - 5 \hat{j} + 0 \hat{k}$ and parallel to $\vec{v}$ is given by $\vec{r} = \vec{a} + \lambda \vec{v}$.
Thus,$\vec{r} = (7 \hat{i} - 5 \hat{j}) + \lambda(3 \hat{i} + 2 \hat{j} - \hat{k})$.

(D) Let $A = (9, 13, 15)$,$B = (12, 21, 10)$,and $P = (5, 7, 3)$. Let $Q(x, y, z)$ be the foot of the perpendicular from $P$ to the line $AB$.
The direction ratios of line $AB$ are $(12 - 9, 21 - 13, 10 - 15) = (3, 8, -5)$.
The equation of line $AB$ is $\frac{x - 9}{3} = \frac{y - 13}{8} = \frac{z - 15}{-5} = \lambda$.
Any point $Q$ on the line $AB$ is given by $Q = (3\lambda + 9, 8\lambda + 13, -5\lambda + 15)$.
The direction ratios of $PQ$ are $(3\lambda + 9 - 5, 8\lambda + 13 - 7, -5\lambda + 15 - 3) = (3\lambda + 4, 8\lambda + 6, -5\lambda + 12)$.
Since $PQ \perp AB$,the dot product of their direction ratios is zero:
$3(3\lambda + 4) + 8(8\lambda + 6) - 5(-5\lambda + 12) = 0$
$9\lambda + 12 + 64\lambda + 48 + 25\lambda - 60 = 0$
$98\lambda = 0 \implies \lambda = 0$.
Substituting $\lambda = 0$ into the coordinates of $Q$,we get $Q = (9, 13, 15)$.
Thus,the foot of the perpendicular is $(9, 13, 15)$,which corresponds to option $D$.

(D) Let the direction ratios of the required line be $(a, b, c)$.
The line passes through the point $P(-1, 3, -2)$.
The direction ratios of the given lines are $\vec{v_1} = (1, 2, 3)$ and $\vec{v_2} = (-3, 2, 5)$.
Since the required line is perpendicular to both given lines,its direction vector $\vec{v} = (a, b, c)$ must be parallel to the cross product $\vec{v_1} \times \vec{v_2}$.
$\vec{v} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & 2 & 5 \end{vmatrix} = \hat{i}(10 - 6) - \hat{j}(5 - (-9)) + \hat{k}(2 - (-6)) = 4\hat{i} - 14\hat{j} + 8\hat{k}$.
Dividing by $2$,we get the direction ratios as $(2, -7, 4)$.
The equation of the line passing through $(-1, 3, -2)$ with direction ratios $(2, -7, 4)$ is $\frac{x - (-1)}{2} = \frac{y - 3}{-7} = \frac{z - (-2)}{4}$,which simplifies to $\frac{x+1}{2} = \frac{y-3}{-7} = \frac{z+2}{4}$.

(A) Let the point of intersection be $P$. The coordinates of any point on the first line are $(1+2\lambda, 2+3\lambda, -1+4\lambda)$ and on the second line are $(-1+\mu, -3+2\mu, 7-\mu)$.
Equating the coordinates,we get:
$1+2\lambda = -1+\mu \implies 2\lambda - \mu = -2$ ... $(i)$
$2+3\lambda = -3+2\mu \implies 3\lambda - 2\mu = -5$ ... $(ii)$
$-1+4\lambda = 7-\mu \implies 4\lambda + \mu = 8$ ... $(iii)$
Adding equations $(i)$ and $(iii)$:
$(2\lambda - \mu) + (4\lambda + \mu) = -2 + 8$
$6\lambda = 6 \implies \lambda = 1$
Substituting $\lambda = 1$ in equation $(i)$:
$2(1) - \mu = -2 \implies \mu = 4$
Checking these values in equation $(ii)$:
$3(1) - 2(4) = 3 - 8 = -5$,which is correct.
Substituting $\lambda = 1$ in the first line equation:
$r = (\hat{i}+2 \hat{j}-\hat{k}) + 1(2 \hat{i}+3 \hat{j}+4 \hat{k}) = 3 \hat{i}+5 \hat{j}+3 \hat{k}$.
Thus,the point of intersection is $3 \hat{i}+5 \hat{j}+3 \hat{k}$.

(A) The shortest distance between two skew lines $r=a+tb$ and $r=c+sd$ is given by the formula: $\text{Shortest distance} = \left| \frac{(c-a) \cdot (b \times d)}{|b \times d|} \right|$.
Given lines are $r=(6 \hat{i}+2 \hat{j}+2 \hat{k})+t(\hat{i}-2 \hat{j}+2 \hat{k})$ and $r=(-4 \hat{i}-\hat{k})+s(3 \hat{i}-2 \hat{j}-2 \hat{k})$.
Here,$a=6 \hat{i}+2 \hat{j}+2 \hat{k}$,$b=\hat{i}-2 \hat{j}+2 \hat{k}$,$c=-4 \hat{i}-\hat{k}$,and $d=3 \hat{i}-2 \hat{j}-2 \hat{k}$.
First,calculate $c-a = (-4 \hat{i}-\hat{k}) - (6 \hat{i}+2 \hat{j}+2 \hat{k}) = -10 \hat{i}-2 \hat{j}-3 \hat{k}$.
Next,calculate the cross product $b \times d = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{vmatrix} = \hat{i}(4+4) - \hat{j}(-2-6) + \hat{k}(-2+6) = 8 \hat{i}+8 \hat{j}+4 \hat{k}$.
The magnitude $|b \times d| = \sqrt{8^2+8^2+4^2} = \sqrt{64+64+16} = \sqrt{144} = 12$.
The dot product $(c-a) \cdot (b \times d) = (-10 \hat{i}-2 \hat{j}-3 \hat{k}) \cdot (8 \hat{i}+8 \hat{j}+4 \hat{k}) = -80 - 16 - 12 = -108$.
Therefore,the shortest distance is $\left| \frac{-108}{12} \right| = |-9| = 9$.

(C) Let the first line be $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$.
Any point on this line is $(2\lambda+1, 3\lambda-1, 4\lambda+1)$.
Since the lines have a point in common,this point must satisfy the second line equation $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$.
Substituting the coordinates into the second line equation:
$\frac{2\lambda+1-3}{1} = \frac{3\lambda-1-k}{2} = \frac{4\lambda+1}{1}$.
Equating the first and third parts:
$2\lambda-2 = 4\lambda+1
\Rightarrow -3 = 2\lambda
\Rightarrow \lambda = -\frac{3}{2}$.
Now,equate the first and second parts:
$\frac{2\lambda-2}{1} = \frac{3\lambda-1-k}{2}$.
Substituting $\lambda = -\frac{3}{2}$:
$2(-\frac{3}{2})-2 = \frac{3(-\frac{3}{2})-1-k}{2}
\Rightarrow -3-2 = \frac{-\frac{9}{2}-1-k}{2}
\Rightarrow -5 = \frac{-\frac{11}{2}-k}{2}
\Rightarrow -10 = -\frac{11}{2}-k
\Rightarrow k = -\frac{11}{2} + 10 = \frac{9}{2}$.

(D) Let the point $P$ be $(3, -2, 1)$. The line passes through $A(1, -3, 5)$ and $B(2, 1, -4)$.
Vector $\vec{a} = \vec{i} - 3\vec{j} + 5\vec{k}$ and the direction vector of the line is $\vec{v} = \vec{AB} = (2-1)\vec{i} + (1-(-3))\vec{j} + (-4-5)\vec{k} = \vec{i} + 4\vec{j} - 9\vec{k}$.
Let $\vec{p} = 3\vec{i} - 2\vec{j} + \vec{k}$. The vector $\vec{AP} = \vec{p} - \vec{a} = (3-1)\vec{i} + (-2-(-3))\vec{j} + (1-5)\vec{k} = 2\vec{i} + \vec{j} - 4\vec{k}$.
The perpendicular distance $d$ is given by $d = \frac{|\vec{AP} \times \vec{v}|}{|\vec{v}|}$.
$\vec{AP} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 1 & -4 \\ 1 & 4 & -9 \end{vmatrix} = \vec{i}(-9 - (-16)) - \vec{j}(-18 - (-4)) + \vec{k}(8 - 1) = 7\vec{i} + 14\vec{j} + 7\vec{k}$.
Magnitude $|\vec{AP} \times \vec{v}| = \sqrt{7^2 + 14^2 + 7^2} = \sqrt{49 + 196 + 49} = \sqrt{294} = 7\sqrt{6}$.
Magnitude $|\vec{v}| = \sqrt{1^2 + 4^2 + (-9)^2} = \sqrt{1 + 16 + 81} = \sqrt{98} = 7\sqrt{2}$.
Distance $d = \frac{7\sqrt{6}}{7\sqrt{2}} = \sqrt{3}$.

(A) 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,$\vec{a_1} = 3\hat{i} + 4\hat{j} - 2\hat{k}$ and $\vec{a_2} = \hat{i} - 7\hat{j} - 2\hat{k}$.
$\vec{a_2} - \vec{a_1} = (1-3)\hat{i} + (-7-4)\hat{j} + (-2 - (-2))\hat{k} = -2\hat{i} - 11\hat{j} + 0\hat{k}$.
$\vec{b_1} = -\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b_2} = \hat{i} + 3\hat{j} + 2\hat{k}$.
$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ 1 & 3 & 2 \end{vmatrix} = \hat{i}(4-3) - \hat{j}(-2-1) + \hat{k}(-3-2) = \hat{i} + 3\hat{j} - 5\hat{k}$.
$|\vec{b_1} \times \vec{b_2}| = \sqrt{1^2 + 3^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$.
$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-2\hat{i} - 11\hat{j} + 0\hat{k}) \cdot (\hat{i} + 3\hat{j} - 5\hat{k}) = (-2)(1) + (-11)(3) + (0)(-5) = -2 - 33 = -35$.
$d = \frac{|-35|}{\sqrt{35}} = \frac{35}{\sqrt{35}} = \sqrt{35}$.

(C) The equation of a line passing through a point $\vec{a} = x_1 \bar{i} + y_1 \bar{j} + z_1 \bar{k}$ and parallel to a vector $\vec{b} = a \bar{i} + b \bar{j} + c \bar{k}$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
Given point is $(1, -2, 1)$,so $x_1 = 1, y_1 = -2, z_1 = 1$.
Given parallel vector is $\bar{i} + \bar{j} + 3 \bar{k}$,so $a = 1, b = 1, c = 3$.
Substituting these values into the formula,we get $\frac{x-1}{1} = \frac{y-(-2)}{1} = \frac{z-1}{3}$.
This simplifies to $\frac{x-1}{1} = \frac{y+2}{1} = \frac{z-1}{3}$.

(A) Let the two lines be $L_1$ and $L_2$. The equation of line $L_1$ is $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$,where $\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
The equation of line $L_2$ is $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$,where $\vec{a}_2 = 2\hat{i} + 4\hat{j} + 5\hat{k}$ and $\vec{b}_2 = 3\hat{i} + 4\hat{j} + 5\hat{k}$.
The shortest distance $d$ between two skew lines 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 $\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$.
Next,calculate $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9) = -\hat{i} + 2\hat{j} - \hat{k}$.
The magnitude $|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{1+4+1} = \sqrt{6}$.
Now,calculate the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (1)(-1) + (2)(2) + (2)(-1) = -1 + 4 - 2 = 1$.
Thus,the shortest distance $d = \frac{|1|}{\sqrt{6}} = \frac{1}{\sqrt{6}}$.

(D) Let the points be $P(2,3,-1)$ and $Q(3,5,-3)$. The direction ratios of the line $PQ$ are $(3-2, 5-3, -3-(-1)) = (1, 2, -2)$.
Let the points be $A(1,2,3)$ and $B(\alpha, \beta, \gamma)$. The direction ratios of the line $AB$ are $(\alpha-1, \beta-2, \gamma-3)$.
Since the lines are perpendicular,the dot product of their direction ratios must be zero:
$1(\alpha-1) + 2(\beta-2) - 2(\gamma-3) = 0$
$\alpha - 1 + 2\beta - 4 - 2\gamma + 6 = 0$
$\alpha + 2\beta - 2\gamma + 1 = 0$.
Now,we check the options for point $B(\alpha, \beta, \gamma)$:
For option $A(-3,5,7)$: $-3 + 2(5) - 2(7) + 1 = -3 + 10 - 14 + 1 = -6 \neq 0$.
For option $B(3,-5,7)$: $3 + 2(-5) - 2(7) + 1 = 3 - 10 - 14 + 1 = -20 \neq 0$.
For option $C(3,5,-7)$: $3 + 2(5) - 2(-7) + 1 = 3 + 10 + 14 + 1 = 28 \neq 0$.
For option $D(3,5,7)$: $3 + 2(5) - 2(7) + 1 = 3 + 10 - 14 + 1 = 0$.
Since the condition is satisfied for option $D$,the correct point is $(3,5,7)$.

(B) The vertices of $\triangle ABC$ are $A(1, 8, 4)$,$B(0, -11, 4)$,and $C(2, -3, 1)$.
First,find the equation of the line $BC$ passing through $B(0, -11, 4)$ and $C(2, -3, 1)$.
The direction ratios of $BC$ are $(2-0, -3-(-11), 1-4) = (2, 8, -3)$.
The equation of line $BC$ is $\frac{x-0}{2} = \frac{y+11}{8} = \frac{z-4}{-3} = \lambda$.
Any point $D$ on the line $BC$ can be represented as $(2\lambda, 8\lambda - 11, -3\lambda + 4)$.
Since $AD \perp BC$,the dot product of vector $\vec{AD}$ and the direction vector of $BC$ must be zero.
Vector $\vec{AD} = (2\lambda - 1, 8\lambda - 11 - 8, -3\lambda + 4 - 4) = (2\lambda - 1, 8\lambda - 19, -3\lambda)$.
The direction vector of $BC$ is $\vec{v} = (2, 8, -3)$.
Setting $\vec{AD} \cdot \vec{v} = 0$:
$2(2\lambda - 1) + 8(8\lambda - 19) + (-3)(-3\lambda) = 0$
$4\lambda - 2 + 64\lambda - 152 + 9\lambda = 0$
$77\lambda - 154 = 0$
$77\lambda = 154 \implies \lambda = 2$.
Substituting $\lambda = 2$ into the coordinates of $D$:
$D = (2(2), 8(2) - 11, -3(2) + 4) = (4, 16 - 11, -6 + 4) = (4, 5, -2)$.

(B) Let the coordinates of the four points $P, Q, R$ and $S$ be $(3, -4, 5), (0, 0, 4), (-4, 5, 1)$ and $(-3, 4, 3)$ respectively.
The equation of line $PQ$ passing through $(3, -4, 5)$ and $(0, 0, 4)$ is given by:
$\frac{x-3}{0-3} = \frac{y+4}{0+4} = \frac{z-5}{4-5} = r_1$
$\Rightarrow \frac{x-3}{-3} = \frac{y+4}{4} = \frac{z-5}{-1} = r_1$
The equation of line $RS$ passing through $(-4, 5, 1)$ and $(-3, 4, 3)$ is given by:
$\frac{x+4}{-3+4} = \frac{y-5}{4-5} = \frac{z-1}{3-1} = r_2$
$\Rightarrow \frac{x+4}{1} = \frac{y-5}{-1} = \frac{z-1}{2} = r_2$
Let a general point on line $PQ$ be $(-3r_1+3, 4r_1-4, -r_1+5)$ and on line $RS$ be $(r_2-4, -r_2+5, 2r_2+1)$.
Since the lines intersect,we equate the coordinates:
$-3r_1+3 = r_2-4 \Rightarrow 3r_1+r_2 = 7$ (Eq. $i$)
$4r_1-4 = -r_2+5 \Rightarrow 4r_1+r_2 = 9$ (Eq. $ii$)
Subtracting Eq. $i$ from Eq. $ii$:
$(4r_1+r_2) - (3r_1+r_2) = 9 - 7 \Rightarrow r_1 = 2$
Substituting $r_1 = 2$ in Eq. $i$:
$3(2) + r_2 = 7 \Rightarrow r_2 = 1$
Substituting $r_2 = 1$ into the coordinates of line $RS$:
$x = 1-4 = -3, y = -1+5 = 4, z = 2(1)+1 = 3$
Thus,the point of intersection is $(-3, 4, 3)$,which corresponds to $-3i+4j+3k$.

(C) The condition for two lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ to be coplanar is $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$.
For the given lines,$(x_1, y_1, z_1) = (3, 2, 1)$ and $(x_2, y_2, z_2) = (2, 3, 2)$.
The direction ratios are $(a_1, b_1, c_1) = (2, 3, \lambda)$ and $(a_2, b_2, c_2) = (3, 2, 3)$.
The vector $\vec{AB} = (2-3, 3-2, 2-1) = (-1, 1, 1)$.
Substituting these into the determinant:
$\begin{vmatrix} -1 & 1 & 1 \\ 2 & 3 & \lambda \\ 3 & 2 & 3 \end{vmatrix} = 0$.
Expanding the determinant: $-1(9-2\lambda) - 1(6-3\lambda) + 1(4-9) = 0$.
$-9 + 2\lambda - 6 + 3\lambda - 5 = 0 \Rightarrow 5\lambda - 20 = 0 \Rightarrow \lambda = 4$.
Now,we need to evaluate $\sin ^{-1}(\sin 4) + \cos ^{-1}(\cos 4)$.
Since $4$ radians is in the third quadrant $(\pi < 4 < \frac{3\pi}{2})$,we use the properties of inverse trigonometric functions:
$\sin ^{-1}(\sin 4) = \sin ^{-1}(\sin(\pi - 4)) = \pi - 4$.
$\cos ^{-1}(\cos 4) = \cos ^{-1}(\cos(2\pi - 4)) = 2\pi - 4$.
Adding these: $(\pi - 4) + (2\pi - 4) = 3\pi - 8$.

(A) Let the $yz$-plane intersect the line segment joining points $A(2, 3, 4)$ and $B(-3, 5, -4)$ in the ratio $\lambda : 1$ at point $M$.
Using the section formula,the coordinates of point $M$ are given by:
$M = \left( \frac{-3\lambda + 2}{\lambda + 1}, \frac{5\lambda + 3}{\lambda + 1}, \frac{-4\lambda + 4}{\lambda + 1} \right)$.
Since the point $M$ lies on the $yz$-plane,its $x$-coordinate must be $0$.
Therefore,$\frac{-3\lambda + 2}{\lambda + 1} = 0$,which implies $-3\lambda + 2 = 0$,so $\lambda = \frac{2}{3}$.
Substituting $\lambda = \frac{2}{3}$ into the coordinates of $M$:
$y = \frac{5(2/3) + 3}{(2/3) + 1} = \frac{10/3 + 9/3}{5/3} = \frac{19}{5}$.
$z = \frac{-4(2/3) + 4}{(2/3) + 1} = \frac{-8/3 + 12/3}{5/3} = \frac{4}{5}$.
Thus,the point of intersection is $\left(0, \frac{19}{5}, \frac{4}{5}\right)$.
Hence,option $A$ is correct.

(C) Line $L_1$ passes through $A(1, 1, 0)$ and $B(-1, 0, 1)$. The direction vector of $L_1$ is $\vec{v_1} = B - A = (-1-1)\hat{i} + (0-1)\hat{j} + (1-0)\hat{k} = -2\hat{i} - \hat{j} + \hat{k}$.
The equation of $L_1$ is $\vec{r} = (1, 1, 0) + s(-2, -1, 1) = (1-2s, 1-s, s)$.
Line $L_2$ passes through $C(0, 1, 2)$ and is parallel to $\vec{v_2} = (1, 1, 1)$.
The equation of $L_2$ is $\vec{r} = (0, 1, 2) + t(1, 1, 1) = (t, 1+t, 2+t)$.
At the intersection point,$(1-2s, 1-s, s) = (t, 1+t, 2+t)$.
Equating coordinates: $1-2s = t$,$1-s = 1+t$,$s = 2+t$.
From $1-s = 1+t$,we get $s = -t$.
Substitute $s = -t$ into $s = 2+t$: $-t = 2+t \implies 2t = -2 \implies t = -1$.
Then $s = 1$.
Intersection point: $x = 1-2(1) = -1$,$y = 1-1 = 0$,$z = 1$.
Check with $L_2$: $x = -1$,$y = 1+(-1) = 0$,$z = 2+(-1) = 1$. Matches.
We need $(y-x) = 0 - (-1) = 1$.
Since $z = 1$,$(y-x) = z$.

(A) The equation of the line $AB$ passing through $A(1, -2, 1)$ and $B(2, -1, 2)$ is given by $\frac{x-1}{2-1} = \frac{y-(-2)}{-1-(-2)} = \frac{z-1}{2-1}$,which simplifies to $\frac{x-1}{1} = \frac{y+2}{1} = \frac{z-1}{1} = K$.
Any point $D$ on the line $AB$ can be represented as $(\alpha, \beta, \gamma) = (K+1, K-2, K+1)$.
The direction ratios of the line $AB$ are $\langle 1, 1, 1 \rangle$.
The vector $\vec{CD} = (\alpha-1, \beta-2, \gamma-3) = (K, K-4, K-2)$.
Since $CD \perp AB$,the dot product of $\vec{CD}$ and the direction vector of $AB$ must be zero:
$1(K) + 1(K-4) + 1(K-2) = 0$.
$3K - 6 = 0 \Rightarrow K = 2$.
Substituting $K=2$,we get $\alpha = 2+1 = 3$,$\beta = 2-2 = 0$,and $\gamma = 2+1 = 3$.
Therefore,$\alpha^2 + \beta^2 + \gamma^2 = 3^2 + 0^2 + 3^2 = 9 + 0 + 9 = 18$.

(B) Since the points $A, B, C, D$ are collinear,they lie on the same line passing through $D(-5, 6, 1)$.
Let the direction ratios of the line be $(p, q, r)$. The equation of the line is $\frac{x+5}{p} = \frac{y-6}{q} = \frac{z-1}{r} = k$.
For point $A(-2, 4, a)$: $\frac{-2+5}{p} = \frac{4-6}{q} = \frac{a-1}{r} \Rightarrow \frac{3}{p} = \frac{-2}{q} = \frac{a-1}{r}$.
For point $B(1, b, 3)$: $\frac{1+5}{p} = \frac{b-6}{q} = \frac{3-1}{r} \Rightarrow \frac{6}{p} = \frac{b-6}{q} = \frac{2}{r}$.
For point $C(c, 0, 4)$: $\frac{c+5}{p} = \frac{0-6}{q} = \frac{4-1}{r} \Rightarrow \frac{c+5}{p} = \frac{-6}{q} = \frac{3}{r}$.
From $\frac{3}{p} = \frac{-2}{q}$ and $\frac{c+5}{p} = \frac{-6}{q}$,we have $\frac{c+5}{3} = \frac{-6}{-2} = 3$ $\Rightarrow c+5 = 9$ $\Rightarrow c = 4$.
From $\frac{6}{p} = \frac{2}{r}$ and $\frac{3}{p} = \frac{a-1}{r}$,we have $\frac{a-1}{3} = \frac{2}{6} = \frac{1}{3}$ $\Rightarrow a-1 = 1$ $\Rightarrow a = 2$.
From $\frac{6}{p} = \frac{b-6}{q}$ and $\frac{3}{p} = \frac{-2}{q}$,we have $\frac{b-6}{3} = \frac{-2}{-2} = 1$ $\Rightarrow b-6 = 3$ $\Rightarrow b = 9$ (Wait,re-evaluating: $\frac{b-6}{q} = \frac{6}{p}$ and $\frac{-2}{q} = \frac{3}{p}$ $\Rightarrow \frac{b-6}{-2} = \frac{6}{3} = 2$ $\Rightarrow b-6 = -4$ $\Rightarrow b = 2$).
Thus,$a=2, b=2, c=4$.
Therefore,$a+b+c = 2+2+4 = 8$.

(B) Let $G_1$ be the centroid of $\triangle ABD$. The coordinates are given by $\left(\frac{1+3-1}{3}, \frac{2+7+0}{3}, \frac{3-2-1}{3}\right) = (1, 3, 0)$.
Let $G_2$ be the centroid of $\triangle ACD$. The coordinates are given by $\left(\frac{1+6-1}{3}, \frac{2+7+0}{3}, \frac{3+7-1}{3}\right) = (2, 3, 3)$.
The position vectors are $\vec{a} = \hat{i} + 3\hat{j}$ and $\vec{b} = 2\hat{i} + 3\hat{j} + 3\hat{k}$.
The vector equation of a line passing through $\vec{a}$ and $\vec{b}$ is $\vec{r} = \vec{a} + t(\vec{b} - \vec{a})$.
$\vec{b} - \vec{a} = (2-1)\hat{i} + (3-3)\hat{j} + (3-0)\hat{k} = \hat{i} + 3\hat{k}$.
Thus,$\vec{r} = (\hat{i} + 3\hat{j}) + t(\hat{i} + 3\hat{k}) = (1+t)\hat{i} + 3\hat{j} + 3t\hat{k}$.

(B) Let $AD$ be the median through vertex $A$. Since $D$ is the midpoint of $BC$,the coordinates of $D$ are given by:
$D = \left(\frac{4+2}{2}, \frac{2+3}{2}, \frac{1+5}{2}\right) = \left(3, \frac{5}{2}, 3\right)$
The vector equation of the line passing through $A(1, 1, 2)$ and $D(3, 5/2, 3)$ is given by $\vec{r} = \vec{a} + t(\vec{d} - \vec{a})$,where $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{d} = 3\hat{i} + \frac{5}{2}\hat{j} + 3\hat{k}$.
$\vec{r} = (\hat{i} + \hat{j} + 2\hat{k}) + t((3-1)\hat{i} + (\frac{5}{2}-1)\hat{j} + (3-2)\hat{k})$
$\vec{r} = (\hat{i} + \hat{j} + 2\hat{k}) + t(2\hat{i} + \frac{3}{2}\hat{j} + \hat{k})$
$\vec{r} = (1+2t)\hat{i} + (1+\frac{3}{2}t)\hat{j} + (2+t)\hat{k}$

(D) Let $D$ be the midpoint of side $AC$. The coordinates of $D$ are given by:
$D = \left( \frac{\alpha + 1}{2}, \frac{5 + 0}{2}, \frac{\beta - 3}{2} \right) = \left( \frac{\alpha + 1}{2}, \frac{5}{2}, \frac{\beta - 3}{2} \right)$
The vector $\vec{BD}$ is given by:
$\vec{BD} = \left( \frac{\alpha + 1}{2} - (-2) \right) \hat{i} + \left( \frac{5}{2} - 1 \right) \hat{j} + \left( \frac{\beta - 3}{2} - 6 \right) \hat{k}$
$\vec{BD} = \left( \frac{\alpha + 5}{2} \right) \hat{i} + \left( \frac{3}{2} \right) \hat{j} + \left( \frac{\beta - 15}{2} \right) \hat{k}$
Since the median $\vec{BD}$ is equally inclined to the coordinate axes,its direction ratios must be equal in magnitude (or proportional). Thus:
$\left| \frac{\alpha + 5}{2} \right| = \left| \frac{3}{2} \right| = \left| \frac{\beta - 15}{2} \right|$
Considering the positive case for the direction ratios:
$\frac{\alpha + 5}{2} = \frac{3}{2} \implies \alpha + 5 = 3 \implies \alpha = -2$
$\frac{\beta - 15}{2} = \frac{3}{2} \implies \beta - 15 = 3 \implies \beta = 18$
Therefore,$\alpha + \beta = -2 + 18 = 16$.

(A) Let the vectors along the two lines be $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = \sqrt{3}\hat{i} - \sqrt{3}\hat{j} + 0\hat{k}$.
First,calculate the magnitudes of the vectors:
$|\vec{a}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.
$|\vec{b}| = \sqrt{(\sqrt{3})^2 + (-\sqrt{3})^2 + 0^2} = \sqrt{3 + 3 + 0} = \sqrt{6}$.
Since the magnitudes are equal $(|\vec{a}| = |\vec{b}|)$,the direction ratios of the angle bisector are given by the components of the vector $\vec{a} + \vec{b}$ or $\vec{a} - \vec{b}$.
Calculating $\vec{a} + \vec{b} = (1 + \sqrt{3})\hat{i} + (1 - \sqrt{3})\hat{j} + (2 + 0)\hat{k}$.
Thus,the direction ratios are $(1 + \sqrt{3}, 1 - \sqrt{3}, 2)$.
Comparing this with the given options,option $A$ is correct.

(A) The given lines are in the form $\vec{r} = \vec{a} + \lambda \vec{b}$.
For the first line,the direction vector is $\vec{b_1} = \hat{i} + 4 \hat{j} + 3 \hat{k}$.
For the second line,the direction vector is $\vec{b_2} = \hat{i} + 2 \hat{j} - 3 \hat{k}$.
The angle $\theta$ between the two lines is given by $\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.
Calculating the dot product: $\vec{b_1} \cdot \vec{b_2} = (1)(1) + (4)(2) + (3)(-3) = 1 + 8 - 9 = 0$.
Since the dot product is $0$,the angle between the lines is $\theta = \cos^{-1}(0) = \frac{\pi}{2}$.

(D) Let the position vector of point $A$ be $\vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$.
The line $L$ passes through $A$ and is parallel to $\vec{v} = 2 \hat{i} + \hat{j} - 2 \hat{k}$.
The equation of the line $L$ is $\vec{r} = \vec{a} + \lambda \vec{v} = (\alpha + 2\lambda) \hat{i} + (\beta + \lambda) \hat{j} + (\gamma - 2\lambda) \hat{k}$.
Point $P$ with position vector $\vec{p} = -7 \hat{i} - 5 \hat{j} + 11 \hat{k}$ lies on $L$,so $\vec{p} = \vec{a} + \lambda \vec{v}$.
Thus,$\vec{p} - \vec{a} = \lambda \vec{v}$,which implies $\overline{AP} = \lambda (2 \hat{i} + \hat{j} - 2 \hat{k})$.
Given $|\overline{AP}| = 12$,we have $|\lambda| |2 \hat{i} + \hat{j} - 2 \hat{k}| = 12$.
Since $|2 \hat{i} + \hat{j} - 2 \hat{k}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4+1+4} = 3$,we get $|\lambda| \times 3 = 12$,so $|\lambda| = 4$,which means $\lambda = \pm 4$.
Since $\vec{a} = \vec{p} - \lambda \vec{v}$,for $\lambda = 4$:
$\vec{a} = (-7 \hat{i} - 5 \hat{j} + 11 \hat{k}) - 4(2 \hat{i} + \hat{j} - 2 \hat{k}) = (-7-8) \hat{i} + (-5-4) \hat{j} + (11+8) \hat{k} = -15 \hat{i} - 9 \hat{j} + 19 \hat{k}$.
For $\lambda = -4$:
$\vec{a} = (-7 \hat{i} - 5 \hat{j} + 11 \hat{k}) + 4(2 \hat{i} + \hat{j} - 2 \hat{k}) = (-7+8) \hat{i} + (-5+4) \hat{j} + (11-8) \hat{k} = \hat{i} - \hat{j} + 3 \hat{k}$.
Comparing with the options,$-15 \hat{i} - 9 \hat{j} + 19 \hat{k}$ is option $D$.

(D) The formula for the shortest distance between two skew lines $\vec{r} = \vec{a_1} + t\vec{b_1}$ and $\vec{r} = \vec{a_2} + s\vec{b_2}$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{ |\vec{b_1} \times \vec{b_2}| }$.
Given $\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{b_1} = \hat{i} + 3\hat{j} + 2\hat{k}$,$\vec{a_2} = 4\hat{i} + 5\hat{j} + 6\hat{k}$,and $\vec{b_2} = 2\hat{i} + 3\hat{j} + \hat{k}$.
First,calculate $\vec{a_2} - \vec{a_1} = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$.
Next,calculate the cross product $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & 2 \\ 2 & 3 & 1 \end{vmatrix} = \hat{i}(3-6) - \hat{j}(1-4) + \hat{k}(3-6) = -3\hat{i} + 3\hat{j} - 3\hat{k}$.
The magnitude $|\vec{b_1} \times \vec{b_2}| = \sqrt{(-3)^2 + 3^2 + (-3)^2} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3}$.
Now,the dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (-3\hat{i} + 3\hat{j} - 3\hat{k}) = -9 + 9 - 9 = -9$.
The shortest distance $d = \frac{|-9|}{3\sqrt{3}} = \frac{9}{3\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.

(D) The $YZ$ plane $(x=0)$ divides $AB$ in ratio $2:3$. Using the section formula,the $x$-coordinate of the division point is $\frac{2(3) + 3(\alpha)}{2+3} = 0$,which gives $6 + 3\alpha = 0$,so $\alpha = -2$.
The $ZX$ plane $(y=0)$ divides $AB$ in ratio $4:5$. Using the section formula,the $y$-coordinate of the division point is $\frac{4(\beta) + 5(4)}{4+5} = 0$,which gives $4\beta + 20 = 0$,so $\beta = -5$.
We need to find the point $C$ that divides $\overline{AB}$ externally in the ratio $\alpha : \beta = -2 : -5$,which is equivalent to $2:5$ externally.
The external division formula for points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ in ratio $m:n$ is $\left(\frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}, \frac{mz_2 - nz_1}{m-n}\right)$.
Substituting $A(-2, 4, 7)$,$B(3, -5, 8)$,$m=2$,and $n=5$:
$x = \frac{2(3) - 5(-2)}{2-5} = \frac{6+10}{-3} = -\frac{16}{3}$.
$y = \frac{2(-5) - 5(4)}{2-5} = \frac{-10-20}{-3} = \frac{-30}{-3} = 10$.
$z = \frac{2(8) - 5(7)}{2-5} = \frac{16-35}{-3} = \frac{-19}{-3} = \frac{19}{3}$.
Thus,the point $C$ is $\left(-\frac{16}{3}, 10, \frac{19}{3}\right)$.

(B) Let $P(2, 3, 4)$ divide the line segment $AB$ in the ratio $k:1$.
Using the section formula,the coordinates of $P$ are given by $\left(\frac{6k+3}{k+1}, \frac{-17k-2}{k+1}, \frac{-4k+2}{k+1}\right)$.
Equating the $x$-coordinate: $\frac{6k+3}{k+1} = 2 \Rightarrow 6k+3 = 2k+2 \Rightarrow 4k = -1 \Rightarrow k = -\frac{1}{4}$.
Since $P$ divides $AB$ internally in the ratio $k:1 = -\frac{1}{4}:1$,the harmonic conjugate $Q$ divides $AB$ externally in the ratio $k:1 = \frac{1}{4}:1$ (or internally in the ratio $\frac{1}{4}:1$).
Using the section formula for $Q$ with ratio $\lambda = \frac{1}{4}$:
$Q = \left(\frac{\frac{1}{4}(6) + 1(3)}{\frac{1}{4}+1}, \frac{\frac{1}{4}(-17) + 1(-2)}{\frac{1}{4}+1}, \frac{\frac{1}{4}(-4) + 1(2)}{\frac{1}{4}+1}\right) = \left(\frac{1.5+3}{1.25}, \frac{-4.25-2}{1.25}, \frac{-1+2}{1.25}\right) = \left(\frac{4.5}{1.25}, \frac{-6.25}{1.25}, \frac{1}{1.25}\right) = \left(\frac{18}{5}, -5, \frac{4}{5}\right)$.
Thus,$\alpha = \frac{18}{5}$,$\beta = -5$,and $\gamma = \frac{4}{5}$.
Therefore,$\alpha + \beta + \gamma = \frac{18}{5} - 5 + \frac{4}{5} = \frac{22}{5} - \frac{25}{5} = -\frac{3}{5}$.

(C) Given vertices of triangle $A(2,3,-1), B(4,1,0)$ and $C(-1,-1,11)$.
First,calculate the lengths of sides $AB$ and $AC$:
$AB = \sqrt{(4-2)^2 + (1-3)^2 + (0+1)^2} = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3$.
$AC = \sqrt{(-1-2)^2 + (-1-3)^2 + (11+1)^2} = \sqrt{(-3)^2 + (-4)^2 + 12^2} = \sqrt{9+16+144} = \sqrt{169} = 13$.
By the Angle Bisector Theorem,the bisector of $\angle BAC$ divides the opposite side $BC$ in the ratio of the sides containing the angle:
$\frac{BD}{DC} = \frac{AB}{AC} = \frac{3}{13}$.
Using the section formula,the coordinates of $D$ are:
$D = \left( \frac{3(-1) + 13(4)}{3+13}, \frac{3(-1) + 13(1)}{3+13}, \frac{3(11) + 13(0)}{3+13} \right) = \left( \frac{-3+52}{16}, \frac{-3+13}{16}, \frac{33+0}{16} \right) = \left( \frac{49}{16}, \frac{10}{16}, \frac{33}{16} \right)$.
The direction ratios of $AD$ are given by the vector $\vec{AD} = D - A$:
$\vec{AD} = \left( \frac{49}{16} - 2, \frac{10}{16} - 3, \frac{33}{16} - (-1) \right) = \left( \frac{49-32}{16}, \frac{10-48}{16}, \frac{33+16}{16} \right) = \left( \frac{17}{16}, \frac{-38}{16}, \frac{49}{16} \right)$.
Since direction ratios can be scaled by a constant,we multiply by $16$ to get $(17, -38, 49)$.
Thus,option $C$ is correct.

(B) The equation of line $L_1$ passing through $(0, -1, 0)$ with direction ratios $(1, 1, 1)$ is given by $\frac{x}{1} = \frac{y+1}{1} = \frac{z}{1} = \lambda$.
Thus,the coordinates of point $A$ are $(\lambda, \lambda-1, \lambda)$.
The line segment $AB$ connects $A(\lambda, \lambda-1, \lambda)$ and $B(1, 1, 1)$.
The direction ratios of line $AB$ are $(\lambda-1, \lambda-2, \lambda-1)$.
Given that the direction ratios of this line are proportional to $2, 1, 2$,we have:
$\frac{\lambda-1}{2} = \frac{\lambda-2}{1} = \frac{\lambda-1}{2}$.
From $\frac{\lambda-1}{2} = \lambda-2$,we get $\lambda-1 = 2\lambda-4$,which implies $\lambda = 3$.
Substituting $\lambda = 3$ into the coordinates of $A$,we get $x = 3$,$y = 3-1 = 2$,and $z = 3$.
Therefore,$x+y+z = 3+2+3 = 8$.

(C) The direction ratios (DRs) of line $AB$ are given by $(x_2-x_1, y_2-y_1, z_2-z_1) = (4-3, 6-4, 3-5) = (1, 2, -2)$.
The direction ratios (DRs) of line $DC$ are given by $(x_4-x_3, y_4-y_3, z_4-z_3) = (1-(-1), 0-2, 5-4) = (2, -2, 1)$.
Let $\theta$ be the angle between the lines $AB$ and $DC$.
The formula for $\cos \theta$ is $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Substituting the values: $\cos \theta = \frac{|(1)(2) + (2)(-2) + (-2)(1)|}{\sqrt{1^2 + 2^2 + (-2)^2} \sqrt{2^2 + (-2)^2 + 1^2}}$.
$\cos \theta = \frac{|2 - 4 - 2|}{\sqrt{1 + 4 + 4} \sqrt{4 + 4 + 1}} = \frac{|-4|}{\sqrt{9} \sqrt{9}} = \frac{4}{3 \times 3} = \frac{4}{9}$.

(C) Given equations are $l+m+n=0$ and $l^2+m^2-n^2=0$.
From the first equation,$l = -(m+n)$.
Substituting this into the second equation: $(-m-n)^2 + m^2 - n^2 = 0$.
$m^2 + 2mn + n^2 + m^2 - n^2 = 0$.
$2m^2 + 2mn = 0 \Rightarrow 2m(m+n) = 0$.
This gives two cases:
Case $1$: $m=0$. Then $l = -n$. The direction ratios are $(-1, 0, 1)$.
Case $2$: $m = -n$. Then $l = 0$. The direction ratios are $(0, -1, 1)$.
Let the two lines be represented by vectors $\vec{a} = -\hat{i} + \hat{k}$ and $\vec{b} = -\hat{j} + \hat{k}$.
The angle $\theta$ between them is given by $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (-1)(0) + (0)(-1) + (1)(1) = 1$.
$|\vec{a}| = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{2}$.
$|\vec{b}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2}$.
$\cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(B) The shortest distance $d$ between two lines $\overline{r} = \overline{a_1} + t\overline{b_1}$ and $\overline{r} = \overline{a_2} + s\overline{b_2}$ is given by the formula $d = \frac{|(\overline{a_2} - \overline{a_1}) \cdot (\overline{b_1} \times \overline{b_2})|}{||\overline{b_1} \times \overline{b_2}||}$.
Here,$\overline{a_1} = 3\bar{i} - 5\bar{j} + 2\bar{k}$,$\overline{b_1} = 4\bar{i} + 3\bar{j} - \bar{k}$,$\overline{a_2} = \bar{i} + 2\bar{j} - 4\bar{k}$,and $\overline{b_2} = 6\bar{i} + 3\bar{j} - 2\bar{k}$.
First,calculate $\overline{a_2} - \overline{a_1} = (1-3)\bar{i} + (2-(-5))\bar{j} + (-4-2)\bar{k} = -2\bar{i} + 7\bar{j} - 6\bar{k}$.
Next,calculate the cross product $\overline{b_1} \times \overline{b_2} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \\ 4 & 3 & -1 \\ 6 & 3 & -2 \end{vmatrix} = \bar{i}(-6 - (-3)) - \bar{j}(-8 - (-6)) + \bar{k}(12 - 18) = -3\bar{i} + 2\bar{j} - 6\bar{k}$.
The magnitude $||\overline{b_1} \times \overline{b_2}|| = \sqrt{(-3)^2 + 2^2 + (-6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.
The dot product $(\overline{a_2} - \overline{a_1}) \cdot (\overline{b_1} \times \overline{b_2}) = (-2)(-3) + (7)(2) + (-6)(-6) = 6 + 14 + 36 = 56$.
Finally,$d = \frac{|56|}{7} = 8$.

(B) First,calculate the lengths of sides $AB$ and $AC$ using the distance formula:
$AB = \sqrt{(1-0)^2 + (5-3)^2 + (6-4)^2} = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$.
$AC = \sqrt{(-2-0)^2 + (0-3)^2 + (-2-4)^2} = \sqrt{(-2)^2 + (-3)^2 + (-6)^2} = \sqrt{4+9+36} = \sqrt{49} = 7$.
By the Angle Bisector Theorem,the bisector of $\angle A$ divides the side $BC$ in the ratio $AB:AC = 3:7$.
Thus,the coordinates of $D$ are given by the section formula:
$D = \left( \frac{3(-2) + 7(1)}{3+7}, \frac{3(0) + 7(5)}{3+7}, \frac{3(-2) + 7(6)}{3+7} \right) = \left( \frac{-6+7}{10}, \frac{0+35}{10}, \frac{-6+42}{10} \right) = \left( \frac{1}{10}, \frac{35}{10}, \frac{36}{10} \right) = (0.1, 3.5, 3.6)$.
Now,calculate the length $AD$:
$AD = \sqrt{(0.1-0)^2 + (3.5-3)^2 + (3.6-4)^2} = \sqrt{(0.1)^2 + (0.5)^2 + (-0.4)^2} = \sqrt{0.01 + 0.25 + 0.16} = \sqrt{0.42} = \sqrt{\frac{42}{100}} = \frac{\sqrt{42}}{10}$.

(B) The line passes through $A(2, -1, 1)$ and $B(1, 1, -2)$. The direction vector of the line is $\vec{v} = (1-2, 1-(-1), -2-1) = (-1, 2, -3)$.
The equation of the line is $\frac{x-2}{-1} = \frac{y+1}{2} = \frac{z-1}{-3} = k$.
Any point on the line is given by $(\alpha, \beta, \gamma) = (2-k, -1+2k, 1-3k)$.
The vector $\vec{PQ} = (\alpha - (-1), \beta - 2, \gamma - (-1)) = (3-k, -3+2k, 2-3k)$ must be perpendicular to the line direction $\vec{v} = (-1, 2, -3)$.
Thus,$\vec{PQ} \cdot \vec{v} = 0 \implies -1(3-k) + 2(-3+2k) - 3(2-3k) = 0$.
$-3 + k - 6 + 4k - 6 + 9k = 0 \implies 14k - 15 = 0 \implies k = \frac{15}{14}$.
Substituting $k$ back: $\alpha = 2 - \frac{15}{14} = \frac{13}{14}$,$\beta = -1 + 2(\frac{15}{14}) = \frac{16}{14}$,$\gamma = 1 - 3(\frac{15}{14}) = -\frac{31}{14}$.
$\alpha + \beta + \gamma = \frac{13+16-31}{14} = -\frac{2}{14} = -\frac{1}{7}$.