Planes in 3D

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Question

If the points $(1, 1, p)$ and $(- 3, 0, 1)$ are equidistant from the plane $r (3 i + 4 j - 12 k) + 13 = 0$ then the value of $3 p$ is $(3 p > 4)$

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Solution

Equation of the plane is
$\begin{array}{l} 3 x + 4 y - 12 z + 13 = 0 \\ | 3 + 4 - 12 p + 13 | = | - 9 - 12 + 13 | \\ \Rightarrow 20 - 12 p = 8 \Rightarrow p = 1, \frac{7}{3} \end{array}$

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