The lines $\frac{x - 3}{1} = \frac{y - 2}{2} = \frac{z - 3}{- l}$ and $\frac{x - 1}{l} = \frac{y - 2}{3} = \frac{z - 1}{4}$ are coplanar if the value of $l$ is

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$2$

$- 11$

$- 13$

$13$

answer is D.

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Detailed Solution

The given lines are $\frac{x - 3}{1} = \frac{y - 2}{2} = \frac{z - 3}{- l}$ and $\frac{x - 1}{l} = \frac{y - 2}{3} = \frac{z - 1}{4}$

The condition that the 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 is $|\begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix}| = 0$

hence,

$\begin{array}{rcl} |\begin{matrix} - 2 & 0 & - 2 \\ 1 & 2 & - l \\ l & 3 & 4 \end{matrix}| & = & 0 \\ - 2 (8 + 3 l) - 2 (3 - 2 l) & = & 0 \\ - 16 - 6 l - 6 + 4 l & = & 0 \\ - 22 - 2 l & = & 0 \\ 2 l & = & - 22 \\ l & = & - 11 \end{array}$