Let the plane $$P$$ pass through the intersection of the planes $$2x + 3y - z = 2$$ and $$x + 2y + 3z = 6$$, and be perpendicular to the plane $$2x + y - z + 1 = 0$$. If $$d$$ is the distance of $$P$$ from the point $$(-7, 1, 1)$$, then $$d^2$$ is equal to:

The plane $$P$$ passes through the intersection of $$2x + 3y - z = 2$$ and $$x + 2y + 3z = 6$$, so the family of planes through this intersection is $$(2 + \lambda)x + (3 + 2\lambda)y + (-1 + 3\lambda)z = 2 + 6\lambda$$.

Because $$P$$ is perpendicular to $$2x + y - z + 1 = 0$$, their normals satisfy $$2(2 + \lambda) + 1(3 + 2\lambda) + (-1)(-1 + 3\lambda) = 0$$. This expands to $$4 + 2\lambda + 3 + 2\lambda + 1 - 3\lambda = 0$$ and hence $$8 + \lambda = 0 \implies \lambda = -8$$.

Substituting $$\lambda = -8$$ yields the equation $$-6x - 13y - 25z = -46$$, which can be rewritten as $$6x + 13y + 25z = 46$$.

The distance from the point $$(-7, 1, 1)$$ to this plane is $$d = \frac{|6(-7) + 13(1) + 25(1) - 46|}{\sqrt{36 + 169 + 625}} = \frac{|-42 + 13 + 25 - 46|}{\sqrt{830}} = \frac{50}{\sqrt{830}},$$ and therefore $$d^2 = \frac{2500}{830} = \boxed{\frac{250}{83}}.$$