Let A (- 2, - 1) , B (1, 0) , C (α, β) and D (γ, ) be the vertices...

Given,

$A$\left(- 2, - 1\right), B\left(1, 0\right), C\left(\alpha, \beta\right)$and$$D\left(\gamma, \delta\right)$ be the vertices of a parallelogram $ABCD$

And the point $C$ lies on $2x - y = 5$ and the point $D$ lies on $3x - 2y = 6$,

Now, plotting the diagram we get,

Now, using the midpoint formula we get,

$P ≡ \left(\dfrac{\alpha - 2}{2}, \dfrac{\beta - 1}{2}\right) ≡ \left(\dfrac{\gamma + 1}{2}, \dfrac{\delta}{2}\right)$

Now, on comparing we get,

$\dfrac{\alpha - 2}{2} = \dfrac{\gamma + 1}{2}$ and $\dfrac{\beta - 1}{2} = \dfrac{\delta}{2}$

$⇒ \alpha - \gamma = 3 . . . ( 1 ) , \beta - \delta = 1 . . . ( 2 )$

Also, given $\left(\gamma, \delta\right)$ lies on $3x - 2y = 6$

$⇒ 3\gamma - 2\delta = 6 . . . \left(3\right)$

And \left(\alpha, \beta\right)$lies on$2x - y = 5$</p><p>$⇒ 2\alpha - \beta = 5 . . . ( 4 )$</p><p>On solving equations$( 1 ) , ( 2 ) , ( 3 ) , ( 4 )$we get,</p><p>$\alpha = - 3, \beta = - 11, \gamma = - 6, \delta = - 12$</p><p>Hence, \left|\alpha + \beta + \gamma + \delta\right| = \left|- 3 - 11 - 6 - 12\right| = 32$