A value of $p$ for which the points $(1,1, p)$ and $(-3,0,1)$ are equidistance from the plane $\vec{r} . (3 \hat{i}+4 \hat{j}-12 \hat{k})+13=0$ is :

$p=1$

Points → A(1, 1, p) and B(-3, 0, 1)

so equation of plane $\vec{r} . (3 \hat{i}+4 \hat{j}-12 \hat{k})+13=0$

let $\vec{r} = (x\hat{i}+y \hat{j} + z \hat{k})0$

⇒  plane is 3x + 4y - 12z + 13 = 0 → equation of a plane

distance of point a 

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

$d_b=\frac{|3(-3)+4(0)-12(1)+13|}{\sqrt{3^2+4^2+(-12)^2}}$

since distance of a point (x₀, y₀, z₀) from plane is $\frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}$

if (ax + by + cz + d = 0) is equation of plane

d_a = d_b

$\Rightarrow \frac{|3(1)+4(1)-12(p)+13|}{\sqrt{4^2+3^2+(-12)^2}} = \frac{|3(-3)+4(0)-12(1)+13|}{\sqrt{4^2+3^2+(-12)^2}}$

$\Rightarrow |3+4-12 p+13|=|-q-12+13|$

$|20-12 p|=|-8|$

$\Rightarrow 4|5-3 p|=4|-2|$

$\Rightarrow|5-3 p|=2$

$\Rightarrow 5-3 p= \pm 2$

so  $3 p=5 \pm 2$

$p=\frac{5 \pm 2}{3}$

$p=1, \frac{7}{3}$