Step 1:
The line $y=mx+c$ is tangent to the ellipse $\large\frac{x^2}{a^2}+\large\frac{y^2}{b^2}$$=1$ if $c=a^2m^2+b^2$
Here the line is $x-y+4=0$ or $y=x+4$.
Therefore $m=1,c=4$
The ellipse is $x^2+3y^2=12$
The above equation is divided by $12$ we get
$\large\frac{x^2}{12}+\large\frac{y^2}{4}$$=1$
Therefore $a^2=12,b^2=4$
Step 2:
We have $a^2m^2+b^2=12\times 1+4=16$
We can see $c^2=16$
Therefore $c^2=a^2m^2+b^2$ and the line is a tangent to the ellipse.
Step 3:
The point of contact is $\big(\large\frac{-a^2m}{c},\large\frac{b^2}{c}\big)$
$\large\frac{-a^2m}{c}=\large\frac{12\times 1}{4}$$=-3$
$\large\frac{b^2}{c}=\large\frac{4}{4}$$=1$
The point of contact is $(-3,1)$