Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

d.r. of given liens are $(1,4,7)$ and $(3,5,7)$. Normal to the plane is perpendicular to them. $$\begin{array}{l}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 4& 7\\ 3& 5& 7\end{array}\right|\\ =\hat{i}(-7)-\hat{j}(-14)+\hat{k}(-7)\\ =-7(\hat{i}-2\hat{j}+\hat{k})\end{array}$$ Hence eq. of plane is $-7(x-2y+z)=a$ This plane plasses through point $(2,4,6)$ $\therefore a=-7(2-8+6)=0$ $\therefore$ Eq. of plane is $x-2y+z=0$