Let the equations of two adjacent sides of a parallelogram $$ABCD$$ be $$2x - 3y = -23$$ and $$5x + 4y = 23$$. If the equation of its one diagonal $$AC$$ is $$3x + 7y = 23$$ and the distance of $$A$$ from the other diagonal is $$d$$, then $$50d^2$$ is equal to _______.

Two adjacent sides of parallelogram ABCD are given by $$2x-3y=-23$$ and $$5x+4y=23$$, while diagonal AC has equation $$3x+7y=23$$. If $$d$$ denotes the distance from A to the other diagonal, we wish to compute $$50d^2$$.

Solving the intersection of the lines $$2x-3y=-23$$ and $$5x+4y=23$$ yields the point $$(-1,7)$$. Since this point satisfies both side‐equations, it must be vertex B. To locate A, we note that A lies on side AB (the line $$2x-3y=-23$$) and on diagonal AC ($$3x+7y=23$$). Solving these simultaneously gives $$A = (-4,5)\,$$.

Similarly, vertex C is the intersection of side BC ($$5x+4y=23$$) and diagonal AC ($$3x+7y=23$$), which leads to $$C = (3,2)\,$$. With A, B, C determined, point D follows from the parallelogram relation $$D = A + (C - B)$$, yielding $$D = (0,0)\,$$. One checks that AD lies on $$5x+4y=0$$ and CD on $$2x-3y=0$$, confirming the parallelogram structure.

The other diagonal BD is the line through $$B = (-1,7)$$ and $$D = (0,0)$$, whose equation can be written as $$7x + y = 0$$. The distance from $$A = (-4,5)$$ to this line is

$$d = \frac{\bigl|7(-4) + 5\bigr|}{\sqrt{7^2 + 1^2}} = \frac{23}{\sqrt{50}}\,. $$

Hence

$$50d^2 = 50 \times \frac{529}{50} = 529\,. $$

The correct answer is 529.