The foot of the perpendicular from $\left(-3,7\right)$  to the line  $3x-4y+12=0$  is

Line $3x-4y+12=0$

$a=3,b=-4,c=12$

Let foot of the ${\perp }^{lar}$  $\left(h,k\right)$

We know that

$\frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-\left(a{x}_1+b{y}_1+c\right)}{a^2+b^2}$

$\Rightarrow \frac{h+3}{3}=\frac{k-7}{-4}=\frac{\left(-3\left(-3\right)-4\left(7\right)+12\right)}{3^2+{\left(-4\right)}^2}$

$\Rightarrow \frac{h+3}{3}=\frac{k-7}{-4}=\frac{-\left(-9-28+12\right)}{25}=\frac{25}{25}=1$

$\frac{h+3}{3}=1$

$h+3=3$

$h=3-3$

$h=0$

$\frac{k-7}{-4}=1$

$k-7=-4$

$k=-4+7=3$

$\therefore$  point $\left(0,3\right)$