Given, area of triangle = 5 sq. units.
Let the third vertex be $(x,y)$.
Area of a triangle formed by $(x_1,y_1),(x_2,y_2)\&(x_3,y_3)=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

$5=\frac{1}{2}|2(-2-y)+3(y-1)+3x|5=\frac{1}{2}|-4-2y+3y-3+3x|$
$3x+y-7=10\text{or}3x+y-7=-10$

$\Rightarrow 3x+y=17\text{or}3x+y=-3$
Also given that the third vertex lies on $y=x+3$. It means that the point of intersection of both the lines is the vertex.
Let us now solve $y=x+3\Rightarrow x-y=-3---\left(1\right)$ and $3x+y=17---\left(2\right)$.
Multiplying $\left(1\right)$ by 3, $\Rightarrow 3x-3y=-9---\left(3\right)$
$\left(2\right)-\left(3\right)\Rightarrow \underset{–}{3x+y=17{}^{}{}_{-}3x\mp 3y=\mp 9}4y=26y=\frac{13}{2}$.
Substituting this value of $y$ in $\left(1\right)$.
$x=y-3=\frac{13}{2}+3=\frac{7}{2}$
$\therefore x=\frac{7}{2}$ and $y=\frac{13}{2}$
 
Now, to solve  $3x-3y=-9---\left(3\right)$ and $3x+y=-3---\left(4\right)$
$\left(4\right)-\left(3\right)\Rightarrow \underset{–}{3x+y=-3{}^{}{}_{-}3x\mp 3y=\mp 9}4y=6y=\frac{3}{2}$.
Substituting this value of $y$ in $\left(1\right)$.
$x=y-3=\frac{3}{2}-3=-\frac{3}{2}$
$\therefore x=-\frac{3}{2}$ and $y=\frac{3}{2}$.
Vertex are $(\frac{7}{2},\frac{13}{2})\&(-\frac{3}{2},\frac{3}{2})$