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(D) The given line is $\frac{x-1}{3} = \frac{y-2}{-m} = \frac{z+3}{1}$. The direction vector of this line is $\vec{v} = (3, -m, 1)$.Let the point $P$

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$2$

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(D) The given line is $\frac{x-1}{3} = \frac{y-2}{-m} = \frac{z+3}{1}$. The direction vector of this line is $\vec{v} = (3, -m, 1)$.Let the point $P$ be $(1, -2, 3)$. $A$ line passing through $P$ parallel to the given line can be written as $\frac{x-1}{3} = \frac{y+2}{-m} = \frac{z-3}{1} = r$.Any point $Q$ on this line is $(1+3r, -2-mr, 3+r)$.Since $Q$ lies on the plane $x + 2y - 3z + 10 = 0$,we have:$(1+3r) + 2(-2-mr) - 3(3+r) + 10 = 0$$1 + 3r - 4 - 2mr - 9 - 3r + 10 = 0$$-2mr - 2 = 0 \Rightarrow -2mr = 2 \Rightarrow mr = -1 \Rightarrow r = -\frac{1}{m}$.The distance $PQ$ is given as $\sqrt{\frac{7}{2}}$.$PQ^2 = (3r)^2 + (-mr)^2 + (r)^2 = r^2(9 + m^2 + 1) = r^2(10 + m^2)$.Substituting $r^2 = \frac{1}{m^2}$:$\frac{7}{2} = \frac{1}{m^2}(10 + m^2)$$\frac{7}{2}m^2 = 10 + m^2$$\frac{5}{2}m^2 = 10 \Rightarrow m^2 = 4 \Rightarrow |m| = 2$.

If the distance of the point $(1, -2, 3)$ from the plane $x + 2y - 3z + 10 = 0$ 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 $|m|$ is equal to ....... .

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