Question:
The distance of the point $(1,1,9)$ from the point
of intersection of the line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$
and the plane $x+y+z=17$ is:
Correct Option: , 4
Solution:
Let $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=t$
$\Rightarrow \quad x=3+t, y=2 t+4, z=2 t+5$
for point of intersection with $x+y+z=17$
$3+t+2 t+4+2 t+5=17$
$\Rightarrow 5 \mathrm{t}=5 \Rightarrow \mathrm{t}=1$
$\Rightarrow$ point of intersection is $(4,6,7)$
distance between $(1,1,9)$ and $(4,6,7)$
is $\sqrt{9+25+4}=\sqrt{38}$