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

Let d.c. of line is $(\mathrm{k},\mathrm{k},\mathrm{k})$ $\begin{array}{l}\Rightarrow 3{\mathrm{k}}^2=1\Rightarrow \mathrm{k}=\frac{1}{\sqrt{3}}\\ \Rightarrow \text{ d.c. of line is }\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)\end{array}$ Equation of a line passing through point $p$ is $\begin{array}{l}\frac{x-2}{\frac{1}{\sqrt{3}}}=\frac{y+1}{\frac{1}{\sqrt{3}}}=\frac{z-2}{\frac{1}{\sqrt{3}}}=k\\ \Rightarrow x=k+2,y=k-1,z=k+2\\ \therefore 2(k+2)+(k-1)+(k+2)=9\\ 4k=4\Rightarrow k=1\end{array}$ $\therefore$ Point $\mathrm{Q}$ is $(3,0,3)$ $PQ=\sqrt{1^2+1^2+1^2}=\sqrt{3}$ units.