A line with positive direction cosines passes through the point \(\mathrm{P}(2,-1,2)\) and makes equal angles with the co-ordinate axes. The line meets the plane \(2 x+y+z=9\) at point \(\mathrm{Q}\). The length of the line segment \(P Q\) equals

Since direction cosines of PQ are equal and positive.
\(\therefore\) The d.r.s. of PQ are \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)
\(\therefore\) The equation of the line \(\mathrm{PQ}\) is
\(
\begin{aligned}
& \frac{x-2}{\frac{1}{\sqrt{3}}}=\frac{y+1}{\frac{1}{\sqrt{3}}}=\frac{\mathrm{z}-2}{\frac{1}{\sqrt{3}}} \\
& \Rightarrow x-2=y+1=\mathrm{z}-2=\mathrm{k}, \text { say }
\end{aligned}
\)
\(\therefore\) Co-ordinates of the point \(\mathrm{Q}\) are
\(
(\mathrm{k}+2, \mathrm{k}-1, \mathrm{k}+2)
\)
The point \(\mathrm{Q}\) lies on the plane \(2 x+y+z=9\)
\(\therefore 2(\mathrm{k}+2)+\mathrm{k}-1+\mathrm{k}+2=9 \)
\( \Rightarrow 4 \mathrm{k}+5=9 \Rightarrow \mathrm{k}=1 \)
\( \therefore \mathrm{Q} \equiv(3,0,3) \)
\( \therefore \mathrm{PQ}=\sqrt{(3-2)^2+(0+1)^2+(3-2)^2} \)
\( =\sqrt{1+1+1}=\sqrt{3}\)