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

Dr 's of $\overrightarrow{\mathrm{AB}}$ = (3, 12, 3)
i.e (1, 4, 1)
Equation of AB: $\frac{\mathrm{x}-4}{1}=\frac{\mathrm{y}-5}{4}=\frac{\mathrm{z}-8}{1}=\mathrm{\lambda }$(say)
$\begin{array}{l}\mathrm{N}=(\mathrm{\lambda }+4,4\mathrm{\lambda }+5,\mathrm{\lambda }+8)\\ {\mathrm{Dr}}^{\mathrm{\prime }}\mathrm{s}\text{ of }\overrightarrow{\mathrm{PN}}=(\mathrm{\lambda }+3,4\mathrm{\lambda }+7,\mathrm{\lambda }+5)\\ \mathrm{PN}\perp \mathrm{AB}\Rightarrow 1(\mathrm{\lambda }+3)+4(4\mathrm{\lambda }+7)+1(\mathrm{\lambda }+5)=0\\ \Rightarrow 18\mathrm{\lambda }+36\Rightarrow \mathrm{\lambda }=-2\\ \mathrm{N}=(2,-3,6)\end{array}$
$\perp \mathrm{r}$ Distance form N to 2x - 2y + z + 5 = 0 is 7.