Let feet of altitudes through $A,B,C$ are $D,E,F$ respectively.
Now, $C$ lies on the line $CF,CA$ & $CB$
Now, the slope of $AD$ is $\frac{7-6}{4-4}=N.D.$
$\Rightarrow$ the slope of $BC$ is $0$ $\Rightarrow$ equation of $BC$ is $y=4$
Now, the slope of $BE$ is $\frac{6-4}{4-(-2)}=\frac{1}{3}$
$\Rightarrow$ the slope of $AC$ is $-3$
$\Rightarrow$ equation of the line $AC$ is $\left(y-7\right)=-3\left(x-4\right)$
Point of intersection of $y=4$ and $y-7=-3\mathrm{}\left(x-4\right)$ gives point $C$

$\Rightarrow C=\left(5,4\right)$