Question:
Let $(\lambda, 2,1)$ be a point on the plane which passes through the ponit $(4,-2,2)$. If the plane is perpendicular to the line joining the points
$(-2,-21,29)$ and $(-1,-16,23)$, then
$\left(\frac{\lambda}{11}\right)^{2}-\frac{4 \lambda}{11}-4$ is equal to
Solution:
$\mid \begin{aligned}&\mathrm{A}(-2,-21,29) \\&\mathrm{B}(-1,-16,33)\end{aligned}$
$\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{PQ}}=0$
$\Rightarrow(\hat{i}+5 \hat{j}-6 \hat{k}) \cdot((4-\lambda) \hat{i}-4 \hat{j}+\hat{k})=0$
$\Rightarrow 4-\lambda-20-6=0$
$\Rightarrow \lambda=-22$
$\Rightarrow\left(\frac{\lambda}{11}\right)^{2}-\frac{4 \lambda}{11}-4=4+8-4=8$