Let $N$ be the foot of perpendicular from the point$P(1,-2,3)$ on the line passing through the points $(4,5,8)$ and $(1,-7,5)$. Then the distance of $N$ from the plane $2x-2y+z+5=0$ is

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,

$\left(4-1,5+7,8-5\right)\equiv \left(3,12,3\right)\equiv \left(1,4,1\right)$

Hence, the equation of Line will be,

$L:\frac{x-1}{1}=\frac{y+7}{4}=\frac{z-5}{1}=r$

Now let point $N\equiv (r+1,4r-7,r+5)$

And given $P\equiv (1,-2,3)$

So, direction ratio of $PN(r, 4r-5, r+2)$

Since, $PN\perp L$ so by perpendicular condition we get,

$r+4(4r-5)+(r+2)=0$

$\Rightarrow r=1$

Hence, $N\equiv (2,-3,6)$

Now finding, distance of $N(2,-3,6)$ from plane $2x-2y+z+5=0$ we get,

$\text{Distance}=\left|\frac{4+6+6+5}{\sqrt{4+4+1}}\right|=7$