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

Let $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=\lambda$ $\therefore$ The co-ordinates of any point on the line are $$\mathrm{P}\equiv (3\lambda +2,4\lambda -1,12\lambda +2)$$ This point lies on the plane $$\begin{array}{rl}& x-y+z=5\\ & \therefore 3\lambda +2-(4\lambda -1)+12\lambda +2=5\\ & \Rightarrow 11\lambda =0\\ & \Rightarrow \lambda =0\\ & \therefore \mathrm{P}\equiv (2,-1,2)\\ & \text{ Let }\mathrm{Q}\equiv (-1,-5,-10)\\ & \therefore \mathrm{PQ}=\sqrt{(-1-2{)}^2+(-5+1{)}^2+(-10-2{)}^2}\\ & =\sqrt{9+16+144}\\ & =13\text{ units }\\ & \end{array}$$