Since, the third vertex $\left(x_{1}, y_{1}\right)$ lie on the line $x+y=5$


$\therefore$\begin{aligned}
x_{1}+y_{1} &=5 \\
y_{1} &=5-x_{1}
\end{aligned}
$\therefore$ Coordinate of $C$ is $\left(x_{1}, 5-x_{1}\right)$
Given, area of $\Delta A B C=4$ units
$\begin{array}{lc}\therefore \frac{1}{2}\left|\begin{array}{ccc}2 & -1 & 1 \\ 3 & 2 & 1 \\ x_{1} & 5-x_{1} & 1\end{array}\right|=4 \\ \text { Using } R_{2} \rightarrow R_{2}-R_{1} \text { and } R_{3} \rightarrow R_{3}-R_{1}, \\ & \left|\begin{array}{ccc}2 & -1 & 1 \\ 1 & 3 & 0 \\ x_{1} & -2 & 6-x_{1} & 0\end{array}\right|=8 \\ \Rightarrow \quad 6-x_{1}-3\left(x_{1}-2\right)=\pm 8 \\ \Rightarrow \quad 6-x_{1}-3 x_{1}+6=\pm 8 \\ \Rightarrow 12-8=4 x_{1} \quad \text { or } \quad 4 x_{1}=20 \\ \Rightarrow \quad x_{1}=1 \quad \quad \text { or } \quad x_{1}=5 \\ \therefore \quad y_{1}=5-1=4 \quad \text { or } \quad y_{1}=0\end{array}$
$\therefore C\left(x_{1}, y_{1}\right)=C(1,4)$ or $C(5,0)$