The given plane is $\vec{r} \cdot (\hat{3i} + \hat{4j} - \hat{12k}) + 13 = 0$ ⇒ 3x + 4y – 12z + 13 = 0 Now given points are equidistant from the given plane $\left| \frac{3 \times 1 + 4 \times 1 - 12p + 13}{\sqrt{9+16+144}} \right|$ $= \left| \frac{3(-3) + 4(0) - 12(1) + 13}{\sqrt{9+16+144}} \right|$⇒ |20 – 12p| = |–8| ⇒ 20 – 12p = ± 8⇒ – 3p = 2 – 5 or – 3p = – 2 – 5 ⇒ – 3p = –3 or – 3p = – 7Hence, $p = 1$ or $p = \frac{7}{3}$