Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the normal vector of the required plane be $\vec{n}$. Since the plane is perpendicular to the given planes, its normal vector is perpendicular to the normal vectors of the given planes.

The normal vectors of the given planes are ${\vec{n}}_1=2\hat{i}+\hat{j}-2\hat{k}$ and ${\vec{n}}_2=3\hat{i}-6\hat{j}-2\hat{k}$.

$\vec{n}={\vec{n}}_1\times {\vec{n}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -2\\ 3& -6& -2\end{array}\right|$

$\vec{n}=\hat{i}(-2-12)-\hat{j}(-4+6)+\hat{k}(-12-3)$

$\vec{n}=-14\hat{i}-2\hat{j}-15\hat{k}$

Taking the direction ratios of the normal as $14,2,15$, the equation of the plane passing through $(1,1,1)$ is:

$14(x-1)+2(y-1)+15(z-1)=0$

$14x+2y+15z=31$

Answer: $14x+2y+15z=31$