Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The d.r.s. of the normal to the plane are $1,-1,2$. $\therefore$ The equation of line PM is $\begin{array}{rl}& \frac{x-2}{1}=\frac{y-4}{-1}=\frac{\mathrm{z}+1}{2}=\lambda (\text{ say })\\ & \Rightarrow x=\lambda +2,y=-\lambda +4,\mathrm{z}=2\lambda -1\\ & \text{ Let }\mathrm{M}\equiv (\lambda +2,-\lambda +4,2\lambda -1)\end{array}$ $\begin{array}{ll}\therefore & \text{ Equation of plane becomes }\\ & 1(\lambda +2)-1(-\lambda +4)+2(2\lambda -1)-2=0\\ & \Rightarrow \lambda =1\\ \therefore & M\equiv (3,3,1)\end{array}$ Since $M$ is the mid-point of $PQ$. $\begin{array}{rl}& \therefore \frac{2+\mathrm{a}}{2}=3,\frac{4+\mathrm{b}}{2}=3,\frac{-1+\mathrm{c}}{2}=1\\ & \Rightarrow \mathrm{a}=4,\mathrm{b}=2,\mathrm{c}=3\\ & \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=9\end{array}$