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

$P(2,-1,2), \quad \text{positive direction cosines}$

$\text{equal angles with the axes } \Rightarrow l = m = n$

$\text{Plane: } 2x+y+z=9$

$\text{To find: length of } PQ$

$l = m = n \quad \text{(equal angles with the axes)}$

$l^{2} + m^{2} + n^{2} = 1 \ \Rightarrow \ 3l^{2} = 1$

$l = m = n = \frac{1}{\sqrt{3}}$

$\text{Direction} \propto \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \propto (1,1,1)$

$L: \frac{x-2}{1} = \frac{y+1}{1} = \frac{z-2}{1}$

$\frac{x-2}{1} = \frac{y+1}{1} = \frac{z-2}{1} = t$

$(x,y,z) = (2+t,\ -1+t,\ 2+t)$

$\text{Plan: } 1.\ \text{put the general point in } 2x+y+z=9$

$2.\ \text{solve for } t \ \Rightarrow \ Q$

$3.\ PQ = \text{distance}(P, Q)$

$2(2+t) + (-1+t) + (2+t) = 9$

$5 + 4t = 9$

$4t = 4 \ \Rightarrow \ t = 1$

$Q = (2+1,\ -1+1,\ 2+1) = (3,0,3)$

$PQ = \sqrt{(3-2)^{2} + (0-(-1))^{2} + (3-2)^{2}}$

$PQ = \sqrt{1 + 1 + 1} = \sqrt{3}$

$PQ = \sqrt{3} \quad \Rightarrow \quad \text{option (3)}$