We have, $A B C D$ is a parallelogram

Let equation of $A B$ be $2 x-3 y=-23 \ldots$ (i)

and equation of $B C$ be $5 x+4 y=23\ldots$ (ii)

Equation of $A C$ is $3 x+7 y=23 \ldots$ (iii)

Solving Eqs. (i) and (ii), we get

$x=-1$, and $y=7$

$\therefore$ Co-ordinate of $B$ is $(-1,7)$

On solving Eqs. (ii) and (iii), we get

$$ x=3, y=2 $$

$\therefore$ Co-ordinate of $C$ is $(3,2)$

On solving Eqs. (i) and (iii), we get $x=-4$ and $y=5$

$\therefore$ Co-ordinate of $A$ is $(-4,5)$.

Let $E$ be the intersection point of diagonal co-ordinate of

$E$ is $\left(\frac{-4+3}{2}, \frac{5+2}{2}\right)$ or $\left(-\frac{1}{2}, \frac{7}{2}\right)$

$\because E$ is mid-point of $A C$

$$ \begin{aligned} & \text { Equation of } B D \text { is } y-7=\left(\frac{7-\frac{7}{2}}{-1+\frac{1}{2}}\right)(x+1) \\\\ & \Rightarrow 7 x+y=0 \end{aligned} $$

Distance of $A$ from diagonal $B D=\frac{|7 \times(-4)+5|}{\sqrt{7^2+1^2}}$

$$ \therefore d=\frac{23}{\sqrt{50}} $$

Hence, $50 d^2=(23)^2=529$