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

Answer 1

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

$\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=\lambda$
$\Rightarrow x=\lambda +3,y=2\lambda +4,z=2\lambda +5$
Which lies on given plane

Hence $\lambda +3+2\lambda +4+2\lambda +5=17$
$\Rightarrow \lambda =1$
Hence, point of intersection is $(4,6,7)$
$\therefore$ Required distance
$=\sqrt{9+25+4}$
$=\sqrt{38}$