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

Equation of line through $Q(1,-2,3)$ and parallel to the line $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$ is $\frac{x-1}{1}=\frac{y+2}{4}=\frac{z-3}{5}=\lambda$ (say) since, point $P$ lies on above line. $\therefore$ $$P(\lambda +1,4\lambda -2,5\lambda +3)$$ since, $P$ lies on the given plane. $\therefore 2(\lambda +1)+3(4\lambda -2)-4(5\lambda +3)+22=0$ $\Rightarrow 2\lambda +2+12\lambda -6-20\lambda -12+22=0$ $$\Rightarrow -6\lambda +6=0$$ $\Rightarrow$ $\lambda =1$ $\therefore$ $P(2,2,8)$ $\therefore PQ=\sqrt{(2-1{)}^2+(2+2{)}^2+(3-8{)}^2}$ $\Rightarrow PQ=\sqrt{1+16+25}=\sqrt{42}$