If the foot of the perpendicular drawn from $$(1, 9, 7)$$ to the line passing through the point $$(3, 2, 1)$$ and parallel to the planes $$x + 2y + z = 0$$ and $$3y - z = 3$$ is $$(\alpha, \beta, \gamma)$$, then $$\alpha + \beta + \gamma$$ is equal to

We need to find $$\alpha + \beta + \gamma$$ where $$(\alpha, \beta, \gamma)$$ is the foot of the perpendicular from $$(1, 9, 7)$$ to the given line.

The line passes through $$(3, 2, 1)$$ and is parallel to both planes $$x + 2y + z = 0$$ and $$3y - z = 3$$. Its direction is the cross product of the normals:

$$\vec{d} = (1, 2, 1) \times (0, 3, -1)$$

$$= \hat{i}(2 \cdot (-1) - 1 \cdot 3) - \hat{j}(1 \cdot (-1) - 1 \cdot 0) + \hat{k}(1 \cdot 3 - 2 \cdot 0)$$

$$= (-5, 1, 3)$$

$$(x, y, z) = (3 - 5t, 2 + t, 1 + 3t)$$

The vector from $$(1, 9, 7)$$ to a point on the line is:

$$\vec{AP} = (2 - 5t, t - 7, 3t - 6)$$

For perpendicularity: $$\vec{AP} \cdot \vec{d} = 0$$:

$$-5(2 - 5t) + 1(t - 7) + 3(3t - 6) = 0$$

$$-10 + 25t + t - 7 + 9t - 18 = 0$$

$$35t - 35 = 0 \implies t = 1$$

$$(\alpha, \beta, \gamma) = (3 - 5, 2 + 1, 1 + 3) = (-2, 3, 4)$$

$$\alpha + \beta + \gamma = -2 + 3 + 4 = 5$$

The correct answer is Option D: $$5$$.