Let N be the foot of perpendicular from the point P ( 1, - 2, 3 )...

Given,

$N$ be the foot of perpendicular from the point$P ( 1, - 2, 3 )$ on the line passing through the points $( 4, 5, 8 ) and ( 1, - 7, 5 )$,

Now finding direction ratio of line $L$ we get,

\left(4 - 1, 5 + 7, 8 - 5\right) ≡ $\left(3, 12, 3\right) ≡ \left(1, 4, 1\right)</p><p>Hence,$ the equation of Line will be,

L: $\dfrac{x - 1}{1} = \dfrac{y + 7}{4} = \dfrac{z - 5}{1} =$r$</p><p>Now let point$N ≡ ( r + 1, 4r - 7, r + 5 )$</p><p>And given$P ≡ ( 1, - 2, 3 )$</p><p>So, direction ratio of$PN ( r, 4r - 5, r + 2 )$</p><p>Since,$PN ⊥ L$so by perpendicular condition we get,</p><p>$r + 4 ( 4r - 5 ) + ( r + 2 ) = 0$</p><p>$⇒ r = 1$</p><p>Hence,$N ≡ ( 2, - 3, 6 )$</p><p>Now finding, distance of$N ( 2, - 3, 6 )$from plane$2x - 2y + z + 5 = 0$we get,</p><p>\text{Distance} =\left|\dfrac{4 + 6 + 6 + 5}{\sqrt{4 + 4 + 1}}\right|$ = 7$</p></blockquote>$