The shortest distance between the lines $\frac{x + 7}{- 6} = \frac{y - 6}{7} = z and \frac{7 - x}{2} = y - 2 = z - 6$ is

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$\sqrt{\frac{37}{29}}$

$2 \sqrt{29}$

$\frac{\sqrt{29}}{2}$

answer is A.

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Detailed Solution

$L_{1} : \frac{x + 7}{- 6} = \frac{y - 6}{7} = \frac{z - 0}{1}$
$Any point on it {\vec{a}}_{1} (- 7, 6, 0)$
$and L_{1} is parallel to {\vec{b}}_{1} (- 6, 7, 1)$
$L_{2} : \frac{x - 7}{- 2} = \frac{y - 2}{1} = \frac{z - 6}{1}$
$Any point on it, a_{2} (7, 2, 6)$
$and L_{2} is parallel to {\vec{b}}_{2} (- 2, 1, 1)$
$Shortest distance between L_{1} and L_{2}$
$\begin{array}{l} = |\frac{({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2})}{|{\vec{b}}_{1} \times {\vec{b}}_{2}|}| = |\frac{(- 14, 4, - 6) \cdot (3, 2, 4)}{\sqrt{9 + 4 + 16}}| \\ = 2 \sqrt{29} . \end{array}$