# The distance of the point

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:

1. (1) $\sqrt{38}$

2. (2) $19 \sqrt{2}$

3. (3) $2 \sqrt{19}$

4. (4) 38

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 $=\mathrm{PQ}$

$=\sqrt{9+25+4}$

$=\sqrt{38}$