Correct option is

(A) $6,-\frac{17}{3}$

Equation of plane $\bar{r} \cdot(3 \hat{i}+2 \hat{j}+6 \hat{k})=13$

i.e. $3 x+2 y+6 z-13=0$

Given point $(2,3, \lambda)$

distance of plane from the points $=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$

$

\begin{array}{l}

5=\left|\frac{3(2)+2(3)+6 \lambda-13}{\sqrt{9+4+36}}\right| \\

\therefore 5=\left|\frac{6 \lambda-1}{7}\right|

\end{array}

$

$\Rightarrow 6 \lambda-1=\pm 35$

$\Rightarrow 6 \lambda=36,6 \lambda=-34$

$

\therefore \lambda=6, \lambda=-\frac{17}{3}

$