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:

• A. $2\sqrt{19}$
• B. $19\sqrt{2}$
• C. $38$
• D. $\sqrt{38}$

Answer 1

The correct option is D $\sqrt{38}$

Let $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=t$
$\Rightarrow x=3+t,y=2t+4,z=2t+5$

For point of intersection with $x+y+z=17$
$3+t+2t+4+2t+5=17$

$\Rightarrow 5t=5\Rightarrow t=1$
$\Rightarrow$ point of intersection is $(4,6,7)$
distance between $(1,1,9)$ and $(4,6,7)$
is $\sqrt{9+25+4}=\sqrt{38}$