Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

$$\text{ Since, }l=m=n=\frac{1}{\sqrt{3}}$$
$\therefore$ Equation of line are $$\begin{array}{c}\frac{x-2}{1/\sqrt{3}}=\frac{y+1}{1/\sqrt{3}}=\frac{z-2}{1/\sqrt{3}}\\ \Rightarrow x-2=y+1=z-2=r\text{ (say) }\end{array}$$ $\therefore$ Any point on the line is $$Q\equiv (r+2r-1,r+2)$$ $\because Q$ lies on the plane $2x+y+z=9$ $$\begin{array}{rl}& \therefore 2(r+2)+(r-1)+(r+2)=9\\ & \Rightarrow 4r+5=9\Rightarrow r=1\Rightarrow Q(3,0,3)\\ & \therefore PQ=\sqrt{(3-2{)}^2+(0+1{)}^2+(3-2{)}^2}\\ & =\sqrt{3}\end{array}$$