If the points $(1,-1, \lambda)$ and $(-3,0,1)$ are equidistant from the plane $3 x-4 y-12 z+13=0$, then the…
If the points $(1,-1, \lambda)$ and $(-3,0,1)$ are equidistant from the plane $3 x-4 y-12 z+13=0$, then the sum of all possible values of $\lambda$ is
$\frac{7}{3}$
$\frac{10}{3}$
$\frac{4}{3}$
$\frac{5}{3}$
Solution
Since the points $(1,-1, \lambda)$ and $(-3,0,1)$ are equidistant from the given plane
$\begin{aligned}
& \left|\frac{3+4-12 \lambda+13}{\sqrt{9+16+144}}\right|=\left|\frac{-9-12+13}{\sqrt{9+16+144}}\right| \\
& \Rightarrow|3+4-12 \lambda+13|=|-9-12+13| \\
& \Rightarrow 20-12 \lambda= \pm 8
\end{aligned}$
$\Rightarrow \lambda=1, \frac{7}{3}$
$\therefore \quad$ Sum of all possible values of $\lambda=1+\frac{7}{3}=\frac{10}{3}$