A circle with centre ( 2, 3 ) and radius 4 intersects the line x +...

A Circle having the centre $\left(2, 3\right)$ and radius $4$.

So, the equation of circle will be

${\left(x - 2\right)}^{2} + {\left(y - 3\right)}^{2} = 16$

$⇒ {x}^{2} + {y}^{2} - 4x - 6y - 3 = 0$

Now, it is given $x + y = 3$ passes through two points $P \text{\text{ and } }$ Q on circle and tangents are drawn at $P \text{\text{ and } }$ Q intersect at $S\left(\alpha, \beta\right)$,

So, $x + y = 3$ will be chord of contact.

Now equation of chord of contact will be

${T}_{\left(\alpha, \beta\right)} = 0$

$⇒ \alphax + \betay - 2\left(x + \alpha\right) - 3\left(y + \beta\right) - 3 = 0$

$⇒ x\left(\alpha - 2\right) + y\left(\beta - 3\right) - 2\alpha - 3\beta - 3 = 0$

Now comparing above equation with $x + y - 3 = 0$, we get

$\dfrac{\alpha - 2}{1} = \dfrac{\beta - 3}{1} = \dfrac{2\alpha + 3\beta + 3}{3}$

Now on solving above relation we get,

$\alpha = - 6 \text{\text{ and } } \beta = - 5$

Hence, the value of

$4\alpha - 7\beta = - 24 + 35 = 11$