Explanation

(b)
Here, $\alpha = \beta = \gamma$
$\because\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1$
$\therefore\ \cos\alpha = \frac{1}{\sqrt{3}}$
DC’s of $P\mathcal{Q}$ are $\left( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right)$
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$PM =$Projection of $AP$ on $P\mathcal{Q}$
$= \left| ( - 2 + 3)\frac{1}{\sqrt{3}} + (3 - 5)\frac{1}{\sqrt{3}} + (1 - 2)\ \frac{1}{\sqrt{3}} \right| = \frac{2}{\sqrt{3}}$
and $AP = \sqrt{( - 2 + 3)^{2} + (3 - 5)^{2} + (1 - 2)^{2}} = \sqrt{6}$
$AM = \sqrt{(AP)^{2} - (PM)^{2}} = \sqrt{6 - \frac{4}{3}} = \sqrt{\frac{14}{3}}$