Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The normal vector of the required plane is perpendicular to the normal vectors of the given planes.

The normal vectors of the given planes are $\overrightarrow{n_1}=\hat{i}+2\hat{j}+2\hat{k}$ and $\overrightarrow{n_2}=3\hat{i}+3\hat{j}+2\hat{k}$.

The normal vector of the required plane is given by their cross product:

$\vec{n}=\overrightarrow{n_1}\times \overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 2\\ 3& 3& 2\end{array}\right|$

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

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

The direction ratios of the normal to the required plane are proportional to $2,-4,3$.

The equation of the plane passing through the point $(1,2,1)$ with normal direction ratios $2,-4,3$ is:

$2(x-1)-4(y-2)+3(z-1)=0$

$2x-2-4y+8+3z-3=0$

$2x-4y+3z+3=0$

Answer: $2x-4y+3z+3=0$