Let the equations of two adjacent sides of a parallelogram ABCD be...

Given,

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

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

Also given, $AC ≡ 3x + 7y = 23$

Solving the above lines we get, A ( – 4, 5 ) , B ( – 1, 7 ) , C ( 3, 2 )$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/490dfafc-db5c-43ed-bf24-a9949cccc74f.png" style="width: 300px; height: 235px;" /></p><p>We know that,</p><p>Diagonal of parallelogram have same midpoint,</p><p>So$AC and BD$have same mid-point and let point$D$be \left(x, y\right),</p><p>So midpoint formula we get, </p><p>\dfrac{x - 1}{2} = \dfrac{- 4 + 3}{2}$ ⇒ x = 0 and $\dfrac{y + 7}{2} = \dfrac{2 + 5}{2} ⇒ y = 0$

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,

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

Hence, $50{d}^{2} = {23}^{2} = 529$