Three Dimensional Geometry — Class 12 Maths Solution

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

$\Rightarrow$ $(x\hat i + y\hat j + z\hat k) \cdot (3\hat i + 4\hat j - 12\hat k) + 13 = 0$

$\Rightarrow$ $3x + 4y - 12z + 13 = 0$

Now, $\left| {\cfrac{{3(1) + 4(1) - 12(p) + 13}}{{\sqrt {9 + 16 + 144} }}} \right| = \left| {\cfrac{{3( - 3) + 4(0) - 12(1) + 13}}{{\sqrt {9 + 16 + 144} }}} \right|$

(Given points are equidistant from the given plane)

$\Rightarrow$ $|20 - 12p| = | - 8| \Rightarrow 20 - 12p = \pm 8 \Rightarrow 5 - 3p = \pm 2$

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

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

or $- 3p = - 7$
Hence, $p = 1$ or $p = \cfrac{7}{3}$