Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Normal to the plane is perpendicular to the given vectors. Hence equation of normal is $$\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 1& -1& 1\\ 3& 2& -1\end{array}\right|=\hat{\mathrm{i}}(-1)-\hat{\mathrm{j}}(-4)+\hat{\mathrm{k}}(5)=-\hat{\mathrm{i}}+4\hat{\mathrm{j}}+5\hat{\mathrm{k}}$$ Hence equation of required plane is $$\begin{array}{rl}& (-1)(\mathrm{x}-2)+(4)(\mathrm{y}-0)+(5)(\mathrm{z}-5)=0\\ & \therefore -\mathrm{x}+2+4\mathrm{y}+5\mathrm{z}-25=0\Rightarrow \mathrm{x}-4\mathrm{y}-5\mathrm{z}+23=0\end{array}$$