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: 1
Solution:
$\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=\lambda$
$\Rightarrow x=\lambda+3, y=2 \lambda+4, z=2 \lambda+5$
Which lines on given plane hence
$\Rightarrow \lambda+3+2 \lambda+4+2 \lambda+5=17$
$\Rightarrow \lambda=\frac{5}{5}=1$
Hence, point of intersection is $\mathrm{Q}(4,6,7)$
$\therefore$ Required distance $=\mathrm{PQ}$
$=\sqrt{9+25+4}$
$=\sqrt{38}$
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