The line passes through $A(-3,4,-8)$ and is parallel to the vector $b =3 \hat{ i }+5 \hat{ j }+6 \hat{ k }$
Let $M$ be the foot of the perpendicular from $P(-2,4,-5)$ on the given line.
We, have
$A P =-\hat{ i }+0 \hat{ j }-3 \hat{ k }$
$\Rightarrow \,| A P |=\sqrt{1+0+9}=\sqrt{10}$
Clearly, AM = Projection of AP on $b$
$\Rightarrow \, A M=\left|\frac{ A P \cdot b }{ b }\right|$
$=\left|\frac{\mid-\hat{ i }-3 \hat{ k }) \cdot(3 \hat{ i }+5 \hat{ j }+6 \hat{ k })}{|3 \hat{ i }+5 \hat{ j }+6 \hat{ k }|}\right|$
$=\left|\frac{-3-18}{\sqrt{9+25+36}}\right|=\left|\frac{-21}{\sqrt{70}}\right|=\frac{21}{\sqrt{70}}$
$\therefore \, P M =\sqrt{A P^{2}-A M^{2}}$
$=\sqrt{10-\frac{441}{70}}=\sqrt{\frac{259}{70}} $
$=\sqrt{\frac{37}{10}} $
