# 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:   Option: 1 $2\sqrt{19}$   Option: 2 $\sqrt{38}$ Option: 3 $38$   Option: 4 $19\sqrt{2}$

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

Which lies on given plane hence

$\\\Rightarrow \lambda+3+2 \lambda+4+2 \lambda+5=17\\\\\Rightarrow \lambda=\frac{5}{5}=1$

\begin{aligned} &\text { Hence, point of intersection is } \mathrm{Q}(4,6,7)\\ &\therefore \text { Required distance }=\mathrm{PQ} \end{aligned}

$PQ=\sqrt{9+25+4} =\sqrt{38}$

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