If the distance of the point $(1, - 2, 3)$ from the plane $x + 2 y - 3 z + 10 = 0$ measured parallel to the line $\frac{x - 1}{3} = \frac{2 - y}{m} = \frac{z + 3}{1}$ is $\sqrt{\frac{7}{2}}$ then the value of $|m| =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$3$

$2$

$1$

$4$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Line passing through $P (1, - 2, 3)$ is $\frac{x - 1}{3} = \frac{y + 2}{- m} = \frac{z - 3}{1} = r$

Foot of perpendicular $Q (3 r + 1, - m r - 2, r + 3)$

Q lies on $\begin{array}{l} x + 2 y - 3 z + 10 = 0 \Rightarrow 3 r + 1 + 2 (- m r - 2) - 3 (r + 3) + 10 = 0 \\ \Rightarrow m r = - 1 \Rightarrow W (3 r + 1, - 1, r + 3) \end{array}$

$\begin{array}{l} P Q^{2} = \frac{7}{2} \Rightarrow {(3 r)}^{2} + 1 + r^{2} = \frac{7}{2} \Rightarrow 10 r^{2} = \frac{5}{2} \Rightarrow r = \frac{1}{2} \\ \Rightarrow m r = - 1 \Rightarrow m (\frac{1}{2}) = - 1 \Rightarrow |m| = 2 \end{array}$