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

1. Direction Cosines: If a line makes equal angles with coordinate axes, its direction ratios are proportional to \((1,1,1)\).
2. Equation of the Line: Parametric form is:
\(\frac{x-2}{1}=\frac{y-1}{1}=\frac{z-2}{1}=t\)
or,
\(x=2+t, y=1+t, z=2+t\)
3. Point of Intersection with Plane: Substituting \(x=2+t, y=1+t, z=2+t\) into \(2 x+\) \(y+z=9\) :
\(\begin{gathered}
2(2+t)+(1+t)+(2+t)=9 \\
4+2 t+1+t+2+t=9 \quad \Rightarrow \quad 4 t+7=9 \\
t=\frac{2}{4}=\frac{1}{2}
\end{gathered}\)
Substituting \(t=\frac{1}{2}\) back, the coordinates of Q are:
\(Q=\left(2+\frac{1}{2}, 1+\frac{1}{2}, 2+\frac{1}{2}\right)=\left(\frac{5}{2}, \frac{3}{2}, \frac{5}{2}\right)\)
4. Length of PQ: Using the distance formula:
\(P Q=\sqrt{\left(\frac{5}{2}-2\right)^2+\left(\frac{3}{2}-1\right)^2+\left(\frac{5}{2}-2\right)^2} \)
\( P Q=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2}=\sqrt{3 \cdot\left(\frac{1}{2}\right)^2}=\) \(\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}\)
Answer: \(2 \sqrt{3}\), Option 2.