Let $$L$$ be a line obtained from the intersection of two planes $$x + 2y + z = 6$$ and $$y + 2z = 4$$. If point $$P(\alpha, \beta, \gamma)$$ is the foot of perpendicular from $$(3, 2, 1)$$ on $$L$$, then the value of $$21(\alpha + \beta + \gamma)$$ equals:

The line $$L$$ is the intersection of the planes $$x + 2y + z = 6$$ and $$y + 2z = 4$$. The direction vector of $$L$$ is the cross product of the normal vectors $$(1, 2, 1)$$ and $$(0, 1, 2)$$, which gives $$(2 \cdot 2 - 1 \cdot 1,\ 1 \cdot 0 - 1 \cdot 2,\ 1 \cdot 1 - 2 \cdot 0) = (3, -2, 1)$$.

To find a point on $$L$$, set $$z = 0$$. Then $$y = 4$$ and $$x + 8 + 0 = 6$$ gives $$x = -2$$. So the point $$(-2, 4, 0)$$ lies on $$L$$.

The parametric equation of $$L$$ is $$(x, y, z) = (-2 + 3t,\ 4 - 2t,\ t)$$.

The vector from a general point on $$L$$ to $$(3, 2, 1)$$ is $$(5 - 3t,\ -2 + 2t,\ 1 - t)$$. For the foot of the perpendicular, this must be orthogonal to the direction vector $$(3, -2, 1)$$, so $$3(5 - 3t) - 2(-2 + 2t) + 1(1 - t) = 0$$.

Expanding: $$15 - 9t + 4 - 4t + 1 - t = 0$$, which gives $$20 - 14t = 0$$, so $$t = \frac{10}{7}$$.

Substituting back: $$\alpha = -2 + \frac{30}{7} = \frac{16}{7}$$, $$\beta = 4 - \frac{20}{7} = \frac{8}{7}$$, $$\gamma = \frac{10}{7}$$.

Therefore, $$\alpha + \beta + \gamma = \frac{16 + 8 + 10}{7} = \frac{34}{7}$$, and $$21(\alpha + \beta + \gamma) = 21 \cdot \frac{34}{7} = 3 \cdot 34 = 102$$.