(Co-ordinate Geometry) Vertices of a Triangle

Question

The area of a triangle is $5$ square units. Two of its vertices are $(2,1)$ and $(3,-2)$. The third vertex lies on the plane $y-x+3=0$.Find the third vertex.

Answer 1

The hint.

Let $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$.

Thus, $$S_{\Delta ABC}=\frac{1}{2}|(x_1-x_2)y_3+(x_2-x_3)y_1+(x_3-x_1)y_2|.$$