A line with positive direction cosines passes through the point \(\mathrm{P}(2,-1,2)\) and makes equal angles with co-ordinate axes. The line meets the plane \(2 x+y+z=9\) at point Q . Then the length of the line segment PQ 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 PQ is
\(\frac{x-2}{\frac{1}{\sqrt{3}}}=\frac{y+1}{\frac{1}{\sqrt{3}}}=\frac{z-2}{\frac{1}{\sqrt{3}}} \)
\(\Rightarrow x-2=y+1=z-2=\mathrm{k}, \text { say}\)
\(\therefore \) Co-ordinates of the point Q are
\((k+2, k-1 ; k+2)\)
The point Q lies on the plane \(2 x+y+z=9\)
\(\therefore 2(k+2)+k-1+k+2=9 \)
\( \Rightarrow 4 k+5=9 \Rightarrow k=1 \)
\( \therefore Q \equiv(3,0,3) \)
\( \therefore P Q=\sqrt{(3-2)^2+(0+1)^2+(3-2)^2} \)
\( =\sqrt{1+1+1} \)
\( =\sqrt{3} \text { units}\)