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(C) The equation of line $L_1$ passing through $A_1 = \hat{i}-2 \hat{j}-\hat{k}$ and $B_1 = 4 \hat{i}-3 \hat{k}$ is given by $r = A_1 + \lambda(B_1 -

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$\frac{4}{3}$

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(C) The equation of line $L_1$ passing through $A_1 = \hat{i}-2 \hat{j}-\hat{k}$ and $B_1 = 4 \hat{i}-3 \hat{k}$ is given by $r = A_1 + \lambda(B_1 - A_1)$.$B_1 - A_1 = (4-1)\hat{i} + (0-(-2))\hat{j} + (-3-(-1))\hat{k} = 3\hat{i} + 2\hat{j} - 2\hat{k}$.So, $L_1: r = (\hat{i}-2 \hat{j}-\hat{k}) + \lambda(3 \hat{i}+2 \hat{j}-2 \hat{k})$.The equation of line $L_2$ passing through $A_2 = \hat{i}+2 \hat{j}-\hat{k}$ and $B_2 = 2 \hat{i}-4 \hat{j}-5 \hat{k}$ is given by $r = A_2 + \mu(B_2 - A_2)$.$B_2 - A_2 = (2-1)\hat{i} + (-4-2)\hat{j} + (-5-(-1))\hat{k} = \hat{i} - 6\hat{j} - 4\hat{k}$.So, $L_2: r = (\hat{i}+2 \hat{j}-\hat{k}) + \mu(\hat{i}-6 \hat{j}-4 \hat{k})$.The shortest distance $D$ between two skew lines $r = a_1 + \lambda b_1$ and $r = a_2 + \mu b_2$ is $D = \frac{|(a_2 - a_1) \cdot (b_1 	imes b_2)|}{|b_1 	imes b_2|}$.$a_2 - a_1 = (\hat{i}+2 \hat{j}-\hat{k}) - (\hat{i}-2 \hat{j}-\hat{k}) = 4\hat{j}$.$b_1 	imes b_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & -2 \ 1 & -6 & -4 \end{vmatrix} = \hat{i}(-8 - 12) - \hat{j}(-12 - (-2)) + \hat{k}(-18 - 2) = -20\hat{i} + 10\hat{j} - 20\hat{k}$.$|b_1 	imes b_2| = \sqrt{(-20)^2 + 10^2 + (-20)^2} = \sqrt{400 + 100 + 400} = \sqrt{900} = 30$.$(a_2 - a_1) \cdot (b_1 	imes b_2) = (4\hat{j}) \cdot (-20\hat{i} + 10\hat{j} - 20\hat{k}) = 4 	imes 10 = 40$.$D = \frac{|40|}{30} = \frac{4}{3}$.

$L_1$ is a line passing through the points with position vectors $\hat{i}-2 \hat{j}-\hat{k}$ and $4 \hat{i}-3 \hat{k}$. $L_2$ is a line passing through the points with position vectors $\hat{i}+2 \hat{j}-\hat{k}$ and $2 \hat{i}-4 \hat{j}-5 \hat{k}$. Then the distance between $L_1$ and $L_2$ is

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