Let the foot of perpendicular of the point $$P(3, -2, -9)$$ on the plane passing through the points $$(-1, -2, -3)$$, $$(9, 3, 4)$$, $$(9, -2, 1)$$ be $$Q(\alpha, \beta, \gamma)$$. Then the distance $$Q$$ from the origin is

$$\overrightarrow{AB} = (9 - (-1))\hat{i} + (3 - (-2))\hat{j} + (4 - (-3))\hat{k} = 10\hat{i} + 5\hat{j} + 7\hat{k}$$

$$\overrightarrow{AC} = (9 - (-1))\hat{i} + (-2 - (-2))\hat{j} + (1 - (-3))\hat{k} = 10\hat{i} + 0\hat{j} + 4\hat{k}$$

$$\vec{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \text{det} \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 10 & 5 & 7 \\ 10 & 0 & 4 \end{bmatrix}$$

$$\vec{n} = \hat{i}(5 \cdot 4 - 7 \cdot 0) - \hat{j}(10 \cdot 4 - 7 \cdot 10) + \hat{k}(10 \cdot 0 - 5 \cdot 10)$$

$$\vec{n} = 20\hat{i} - \hat{j}(40 - 70) - 50\hat{k} = 20\hat{i} + 30\hat{j} - 50\hat{k}$$

$$\vec{n}_{\text{scaled}} = 2\hat{i} + 3\hat{j} - 5\hat{k}$$

Using point $$C(9, -2, 1)$$, the scalar equation of the plane is:

$$2(x - 9) + 3(y - (-2)) - 5(z - 1) = 0$$

$$2x - 18 + 3y + 6 - 5z + 5 = 0$$

$$2x + 3y - 5z - 7 = 0$$

Using the foot of the perpendicular formula:

$$\frac{\alpha - 3}{2} = \frac{\beta - (-2)}{3} = \frac{\gamma - (-9)}{-5} = -\frac{2(3) + 3(-2) - 5(-9) - 7}{2^2 + 3^2 + (-5)^2}$$

$$\text{Ratio} = -\frac{38}{38} = -1$$

$$\frac{\alpha - 3}{2} = -1 \implies \alpha - 3 = -2 \implies \alpha = 1$$

$$\frac{\beta + 2}{3} = -1 \implies \beta + 2 = -3 \implies \beta = -5$$

$$\frac{\gamma + 9}{-5} = -1 \implies \gamma + 9 = 5 \implies \gamma = -4$$

The coordinates of the foot of the perpendicular are $$Q(1, -5, -4)$$

$$OQ = \sqrt{1^2 + (-5)^2 + (-4)^2}$$

$$OQ = \sqrt{1 + 25 + 16} = \sqrt{42}$$