Let $$P_1$$ be the plane $$3x - y - 7z = 11$$ and $$P_2$$ be the plane passing through the points $$(2, -1, 0)$$, $$(2, 0, -1)$$, and $$(5, 1, 1)$$. If the foot of the perpendicular drawn from the point $$(7, 4, -1)$$ on the line of intersection of the planes $$P_1$$ and $$P_2$$ is $$(\alpha, \beta, \gamma)$$, then $$\alpha + \beta + \gamma$$ is equal to _____.

$$P_1: 3x - y - 7z = 11$$

$$P_2$$ passes through $$(2,-1,0), (2,0,-1), (5,1,1)$$.

Vectors in P₂: $$(2,0,-1)-(2,-1,0) = (0,1,-1)$$ and $$(5,1,1)-(2,-1,0) = (3,2,1)$$.

Normal to P₂: $$(0,1,-1) \times (3,2,1) = (1+2, -(0+3), (0-3)) = (3, -3, -3) = 3(1,-1,-1)$$.

$$P_2: 1(x-2) - 1(y+1) - 1(z-0) = 0$$

$$x - y - z = 3$$

Line of intersection of $$P_1$$ and $$P_2$$:

Direction = $$\vec{n_1} \times \vec{n_2} = (3,-1,-7) \times (1,-1,-1)$$

$$= (1-7, -(-3+7), (-3+1)) = (-6, -4, -2) = -2(3, 2, 1)$$

Finding a point on both planes: Let $$z = 0$$:

$$3x - y = 11$$ and $$x - y = 3$$. Subtracting: $$2x = 8$$, $$x = 4$$, $$y = 1$$.

Point: $$(4, 1, 0)$$.

Line: $$\frac{x-4}{3} = \frac{y-1}{2} = \frac{z}{1} = t$$

Foot of perpendicular from $$(7,4,-1)$$ to this line:

Point on line: $$(4+3t, 1+2t, t)$$

Vector from point to $$(7,4,-1)$$: $$(3-3t, 3-2t, -1-t)$$

This should be perpendicular to direction $$(3,2,1)$$:

$$3(3-3t) + 2(3-2t) + 1(-1-t) = 0$$

$$9 - 9t + 6 - 4t - 1 - t = 0$$

$$14 - 14t = 0$$

$$t = 1$$

Foot = $$(7, 3, 1)$$. So $$\alpha + \beta + \gamma = 7 + 3 + 1 = 11$$.

The answer is 11. ✓