Given,

The equations of two adjacent sides of a parallelogram $ABCD$ be $2x-3y=-23$ and $5x+4y=23$,

So, $AB\equiv 2x-3y=-23$ and $BC\equiv 5x+4y=23$

Also given, $AC\equiv 3x+7y=23$

Solving the above lines we get, $A(–4, 5), B(–1, 7), C(3, 2)$

We know that,

Diagonal of parallelogram have same midpoint,

So $AC$ and $BD$ have same mid-point and let point $D$ be $\left(x,y\right)$,

So midpoint formula we get, 

$\frac{x-1}{2}=\frac{-4+3}{2}\Rightarrow x=0$ and $\frac{y+7}{2}=\frac{2+5}{2}\Rightarrow y=0$

Hence, point $D$ is $(0, 0)$

Now Equation of $BD$ will be $7x+y=0$

Now finding the distance of $A\left(-4,5\right)$ from $7x+y=0$ we get,

$d=\left|\frac{7(-4)+5}{\sqrt{7^2+1^2}}\right|=\frac{23}{\sqrt{50}}$

Hence, $50d^2={23}^2=529$