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The distance of the point (4, 6, -2) from the line passing through the point (-3, 2, 3) and parallel to a line with direction ratios 3, 3, -1 is equal to :

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$\sqrt{6}$

$2 \sqrt{3}$

$\sqrt{14}$

answer is B.

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Detailed Solution

$\begin{array}{l} d = 3 \sqrt{10} \\ cosθ = \frac{21 + 12 + 5}{3 \sqrt{10} . \sqrt{19}} = \frac{2 \sqrt{19}}{3 \sqrt{10}} \\ = \frac{\sqrt{38}}{\sqrt{45}} ∴ sinθ = \frac{\sqrt{7}}{\sqrt{45}} \\ ∴ d sinθ = \sqrt{14} \end{array}$

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