Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The plane contains the line $\frac{x-2}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$, so it passes through the point $B(2,3,-2)$ and is parallel to the vector $\vec{b}=-3\hat{i}+2\hat{j}+\hat{k}$.

The plane also passes through the point $A(3,7,-7)$.

The vector joining $A$ and $B$ is $\overrightarrow{AB}=(2-3)\hat{i}+(3-7)\hat{j}+(-2-(-7))\hat{k}=-\hat{i}-4\hat{j}+5\hat{k}$.

The normal vector $\vec{n}$ to the plane is perpendicular to both $\vec{b}$ and $\overrightarrow{AB}$.

$\vec{n}=\vec{b}\times \overrightarrow{AB}=\left|\begin{array}{ccccccc}\hat{i}& \hat{j}& \hat{k}-3& 2& 1-1& -4& 5\end{array}\right|$

$\vec{n}=\hat{i}(10+4)-\hat{j}(-15+1)+\hat{k}(12+2)=14\hat{i}+14\hat{j}+14\hat{k}$.

The direction ratios of the normal are proportional to $1,1,1$.

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

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

$x+y+z-3=0$

The distance $d$ of the plane from the origin $(0,0,0)$ is:

$d=\frac{|0+0+0-3|}{\sqrt{1^2+1^2+1^2}}=\frac{3}{\sqrt{3}}=\sqrt{3}$

$d^2=3$