Let $${{x + 3} \over {10}}$$= $${{y - 2} \over {-7}}$$ = $${{z} \over {1}}$$ = $$\lambda $$

$$ \therefore $$ General point Q(10$$\lambda $$ - 3, -7$$\lambda $$ + 2, $$\lambda $$).

$$\overrightarrow {PQ} $$ = (10$$\lambda $$ - 5)$$\widehat i$$ + (3 - 7$$\lambda $$)$$\widehat j$$ + ($$\lambda $$ - 4)$$\widehat k$$

The vector parrallel to the given line is

= 10$$\widehat i$$ - 7$$\widehat j$$ + $$\widehat k$$

As $$\overrightarrow {PQ} $$ and (10$$\widehat i$$ - 7$$\widehat j$$ + $$\widehat k$$) are perpendicular to the each other.

$$\overrightarrow {PQ} $$ . (10$$\widehat i$$ - 7$$\widehat j$$ + $$\widehat k$$) = 0

$$ \Rightarrow $$ (10$$\lambda $$ - 5)10 + (3 - 7$$\lambda $$)(-7) + ($$\lambda $$ - 4)(1) = 0

$$ \Rightarrow $$ 100$$\lambda $$ – 50 + 49$$\lambda $$ – 21 + $$\lambda $$ – 4 = 0

$$ \Rightarrow $$ 150$$\lambda $$ = 75

$$ \Rightarrow $$ $$\lambda $$ = $${1 \over 2}$$

$$ \therefore $$ The point Q = (2, $$ - {3 \over 2}$$, $${1 \over 2}$$)

$$ \therefore $$ Length of perpendicular (PQ) = $$\sqrt {0 + {1 \over 4} + {{49} \over 4}} $$

= $$\sqrt {{{50} \over 4}} = {5 \over {\sqrt 2 }}$$ = 3.53