# If the points $(1,1,p)$ and $(-3,0,1)$ are at equidistant from the plane $\overrightarrow {r} . (3 \hat {i} +4 \hat {j} -12 \hat {k}) + 13 = 0,$ then find the value of p.

$(a)\; 1 , \; 7/3$
$(b)\; -1, \; -7/3$
$(c)\; -1, \; 7/3$
$(d)\; 1, \;-7/3$

The distance of point $(1, 1, p)$ from the given plane is
\begin{align*}d_1 &= \begin{vmatrix} \frac{3 \times 1 + 4 \times 1 - 12 \times p + 13}{3^2 +4^2+(-12)^2} \end{vmatrix} \\ & = \begin{vmatrix} \frac{20-12p}{13} \end{vmatrix} \end{align*}
\begin{align*} d_2 & =\begin{vmatrix} \frac{3 \times -3 + 4 \times 0 - 12 \times 1 +13}{\sqrt{3^2 +4^2 +(-12)^2}} \end{vmatrix} \\ & = \frac{-8}{13} \end{align*}
\begin{align*} Since \; d_1 = d_2, \begin{vmatrix} \frac{20 - 12p}{13} \end{vmatrix}& = \pm \frac{8}{13} \\ \implies \frac{20 -12p}{13} & = \pm \frac{8}{13} \\ \implies 12 p & = 12 \\ \therefore p & = 1 \\ and \; 12 p & = 28 \\ \therefore p &= 7/3\end{align*}