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1. ⇒ (MHT CET 2023 12th May Morning Shift )

The distance of the point $P (- 2, 4, - 5)$ from the line $\frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6}$ is

A. $\frac{\sqrt{37}}{10}$

B. $\sqrt{\frac{37}{10}}$

C. $\frac{37}{\sqrt{10}}$

D. $\frac{37}{10}$

Correct answer option is (B)

Since the point is $(- 2, 4, - 5)$ ,

$∴ a = - 2, b = 4, c = - 5$

Given equation of line is

$\frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6}$

$x_{1} = - 3, y_{1} = 4, z_{1} = - 8$

d.r.s of the line are $3, 5, 6$

d.c.s are $\frac{3}{\sqrt{70}}, \frac{5}{\sqrt{70}}, \frac{6}{\sqrt{70}}$

Perpendicular distance of point from the line is

$\begin{aligned} \sqrt{\begin{array}{r} [{(a - x_{1})}^{2} + {(b - y_{1})}^{2} + {(c - z_{1})}^{2}] \\ - {[(a - x_{1}) l + (b - y_{1}) m + (c - z_{1}) n]}^{2} \end{array}} \\ = \sqrt{1^{2} + 0 + 3^{2} - {[\frac{1 (3)}{\sqrt{70}} + \frac{0 (5)}{\sqrt{70}} + \frac{3 (6)}{\sqrt{70}}]}^{2}} \\ = \sqrt{1 + 9 - {(\frac{3}{\sqrt{70}} + \frac{18}{\sqrt{70}})}^{2}} \\ = \sqrt{\frac{37}{10}} .units \end{aligned}$

⇒Line and Plane