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 $$x + 2y + 3z = 6$$ and is perpendicular to $$2x + y - z + 1 = 0$$. Any plane through the intersection of the two given planes can be written as $$(2x + 3y - z - 2) + \lambda(x + 2y + 3z - 6) = 0$$ which simplifies to $$(2 + \lambda)x + (3 + 2\lambda)y + (-1 + 3\lambda)z - (2 + 6\lambda) = 0$$.

The normal to plane $$P$$ is $$\vec{n_1} = (2+\lambda, 3+2\lambda, -1+3\lambda)$$, and the normal to $$2x + y - z + 1 = 0$$ is $$\vec{n_2} = (2, 1, -1)$$. For perpendicularity we require $$\vec{n_1} \cdot \vec{n_2} = 0$$, i.e. $$2(2+\lambda) + (3+2\lambda) + (-1)(-1+3\lambda) = 0$$ which gives $$4 + 2\lambda + 3 + 2\lambda + 1 - 3\lambda = 0$$ and hence $$8 + \lambda = 0$$ so $$\lambda = -8$$.

Substituting $$\lambda = -8$$ into the plane equation yields $$(2-8)x + (3-16)y + (-1-24)z - (2-48) = 0$$, i.e. $$-6x - 13y - 25z + 46 = 0$$, or equivalently $$6x + 13y + 25z = 46$$.

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

The correct answer is Option (1): $$\boxed{\frac{250}{83}}$$.