Answer:
Option A
Explanation:
Given point be (2,3,$\lambda$) and equation of the plane be
$r.(3\hat{i}+2\hat{j}+6\hat{k})=13$
$\therefore$ $ (x\hat{i}+y\hat{j}+z\hat{k}).(3\hat{i}+2\hat{j}+6\hat{k})=13$
$ [\because r.(x\hat{i}+y\hat{j}+z\hat{k})]$
$\Rightarrow $ $ 3x+2y+6z=13$
or 3x+2y+6z-13=0
Now, distance of the plane from the point (2,3, $\lambda$)
$d=|\frac{ax_{1}+by_{1}+cz_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}|$
$5=|\frac{3\times2+2\times3+6 \times\lambda-13}{\sqrt{3^{2}+2^{2}+6^{2}}}|$
[given , d=5]
$\pm5=\frac{6+6+6 \times\lambda-13}{\sqrt{9+4+36}}$
$\Rightarrow $ $\pm5=\frac{6\lambda-1}{\sqrt{49}}$
$\therefore$ $5=|\frac{6\lambda-1}{7}|$
$\Rightarrow$ $\pm35=6\lambda-1$
$\Rightarrow$ $35=6\lambda-1$
or $-35=6\lambda-1$
$\Rightarrow$ $\lambda=6,\lambda=-\frac{17}{3}$
