Let a plane P pass through the point $(3,7,-7)$ and
contain the line, $\frac{x-2}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$. If distance
of the plane $P$ from the origin is $d$, then $d^{2}$ is equal to
$\overrightarrow{\mathrm{BA}}=(\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})$
$\overrightarrow{\mathrm{BA}}=(\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})$
$\overrightarrow{\mathrm{BA}} \times \vec{\ell}=\overrightarrow{\mathrm{n}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -3 & 2 & 1 \\ 1 & 4 & -5\end{array}\right|$
$a \hat{i}+b \hat{j}+c \hat{k}=-14 \hat{i}-\hat{j}(14)+\hat{k}(-14)$
$\mathrm{a}=1, \mathrm{~b}=1, \mathrm{c}=1$
Plane is $(x-2)+(y-3)+(z+2)=0$
$x+y+z-3=0$
$\mathrm{d}=\sqrt{3} \Rightarrow \mathrm{d}^{2}=3$