We know that, equation of a plane passing through 3 points, $\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix}$ = 0 ⇒ $\begin{vmatrix} x-3 & y+1 & z-2 \\ 2 & 3 & 2 \\ -4 & 0 & 4 \end{vmatrix}$ = 0 ⇒ (x - 3) (12 - 0) - (y + 1) (8 + 8) + (z - 2) (0 + 12) = 0 ⇒ 12x - 36 - 16y - 16 + 12z - 24 = 0 ⇒ 12x - 16y + 12z - 76 = 0 ⇒ 3x - 4y + 3z - 19 = 0 Also ,perpendicular distance of P(6, 5, 9) to the plane 3x - 4y + 3z - 19 = 0 = $\frac{\left| 3 \times 6 - 4 \times 5 + 3 \times 9 - 19 \right|}{\sqrt{9+16+9}}$ = $\frac{6}{\sqrt{34}}$ units