If the points

Solution:

The position vector through the point $(1,1, p)$ is $\vec{a}_{1}=\hat{i}+\hat{j}+p \hat{k}$

Similarly, the position vector through the point $(-3,0,1)$ is

$\vec{a}_{2}=-4 \hat{i}+\hat{k}$

The equation of the given plane is $\vec{r} \cdot(3 \hat{i}+4 \hat{j}-12 \hat{k})+13=0$

It is known that the perpendicular distance between a point whose position vector is $\vec{a}$ and the plane, $\vec{r} \cdot \vec{N}=d$, is given by, $D=\frac{|\vec{a} \cdot \vec{N}-d|}{|\vec{N}|}$

Here, $\vec{N}=3 \hat{i}+4 \hat{j}-12 \hat{k}$ and $d=-13$

Therefore, the distance between the point (1, 1, p) and the given plane is

$D_{1}=\frac{|(\hat{i}+\hat{j}+p \hat{k}) \cdot(3 \hat{i}+4 \hat{j}-12 \hat{k})+13|}{|3 \hat{i}+4 \hat{j}-12 \hat{k}|}$

$\Rightarrow D_{1}=\frac{|3+4-12 p+13|}{\sqrt{3^{2}+4^{2}+(-12)^{2}}}$

$\Rightarrow D_{1}=\frac{|20-12 p|}{13}$                              $\ldots(1)$

Similarly, the distance between the point (−3, 0, 1) and the given plane is

$D_{2}=\frac{|(-3 \hat{i}+\hat{k}) \cdot(3 \hat{i}+4 \hat{j}-12 \hat{k})+13|}{|3 \hat{i}+4 \hat{j}-12 \hat{k}|}$

$\Rightarrow D_{2}=\frac{|-9-12+13|}{\sqrt{3^{2}+4^{2}+(-12)^{2}}}$

$\Rightarrow D_{2}=\frac{8}{13}$                         ...(3)

It is given that the distance between the required plane and the points, (1, 1, p) and (−3, 0, 1), is equal.

∴ D1 = D2

$\Rightarrow \frac{|20-12 p|}{13}=\frac{8}{13}$

$\Rightarrow 20-12 p=8$ or $-(20-12 p)=8$

$\Rightarrow 12 p=12$ or $12 p=28$

$\Rightarrow p=1$ or $p=\frac{7}{3}$