Let P be the plane, passing through the point (1, - 1, - 5) and...

Given,

$P$ be the plane, passing through the point \left(1, - 1, - 5\right) and perpendicular to the line joining the points \left(4, 1, - 3\right) and \left(2, 4, 3\right),

So, direction ratio of Plane Normal will be, \left(4 - 2, 1 - 4, - 3 - 3\right) ≡ \left(2, - 3, - 6\right)

Now, equation of Plane will be,

$2\left(x - 1\right)$ - 3 \left(y + 1\right)$- 6 \left(z + 5\right)$ = 0$</p><p>$⇒ 2x - 3y - 6z - 35 = 0

Hence, perpendicular distance of $plane from \left(3, - 2, 2\right) will be,</p><p>Distance$= \left|\dfrac{2 \times 3 - 3 \times $\left(- 2\right)$ - 6 \times 2 - 35}{\sqrt{{2}^{2} + {\left(- 3\right)}^{2} + {\left(- 6\right)}^{2}}}\right| = \left|\dfrac{6 + 6 - 12 - 35}{\sqrt{{2}^{2} + {\left(- 3\right)}^{2} + {\left(- 6\right)}^{2}}}\right| = 5$.</p>$