If the foot of the perpendicular from $\left(-4,5\right)$   to the straight line $3x-4y-18=0$  is  $\left(\alpha ,\beta \right)$ then the value of $\alpha +\beta$ is

Line $3x-4y-18=0$

$a=3,b=-4,c=-18$

$\left(\alpha ,\beta \right)$  is foot of the perpendicular

So, $\frac{\alpha -x_1}{a}=\frac{\beta -y_1}{b}=\frac{-\left(a{x}_1-b{y}_1+c\right)}{a^2+b^2}$

$\Rightarrow \frac{\alpha +4}{3}=\frac{\beta -5}{-4}=\frac{-\left(-12-20-18\right)}{25}$

$\Rightarrow \frac{\alpha +4}{3}=\frac{\beta -5}{-4}=\frac{50}{25}=2$

$\frac{\alpha +4}{3}=2$

$\alpha +4=6$

$\alpha =6-4=2$

$\frac{\beta -5}{-4}=2$

$\beta -5=-8$

$\beta =-8+5=-3$

$\therefore$   $\alpha +\beta =2-3=-1$