Determine,algebraically,the vertices of the triangle formed by the lines:
$3x - y = 3$
$2x - 3y = 2$
$x + 2y = 8$

(A) The given equations of the lines are:
$3x - y = 3 \quad ...(i)$
$2x - 3y = 2 \quad ...(ii)$
$x + 2y = 8 \quad ...(iii)$
Let the lines $(i), (ii),$ and $(iii)$ represent the sides $AB, BC,$ and $CA$ of $\triangle ABC$ respectively.
$1$. To find vertex $B$,solve lines $(i)$ and $(ii)$:
Multiply Eq. $(i)$ by $3$: $9x - 3y = 9$
Subtract Eq. $(ii)$ from this: $(9x - 3y) - (2x - 3y) = 9 - 2 \Rightarrow 7x = 7 \Rightarrow x = 1$.
Substitute $x = 1$ in Eq. $(i)$: $3(1) - y = 3 \Rightarrow y = 0$.
So,vertex $B$ is $(1, 0)$.
$2$. To find vertex $C$,solve lines $(ii)$ and $(iii)$:
Multiply Eq. $(iii)$ by $2$: $2x + 4y = 16$.
Subtract Eq. $(ii)$ from this: $(2x + 4y) - (2x - 3y) = 16 - 2 \Rightarrow 7y = 14 \Rightarrow y = 2$.
Substitute $y = 2$ in Eq. $(iii)$: $x + 2(2) = 8 \Rightarrow x = 4$.
So,vertex $C$ is $(4, 2)$.
$3$. To find vertex $A$,solve lines $(iii)$ and $(i)$:
Multiply Eq. $(i)$ by $2$: $6x - 2y = 6$.
Add Eq. $(iii)$: $(6x - 2y) + (x + 2y) = 6 + 8 \Rightarrow 7x = 14 \Rightarrow x = 2$.
Substitute $x = 2$ in Eq. $(i)$: $3(2) - y = 3 \Rightarrow 6 - y = 3 \Rightarrow y = 3$.
So,vertex $A$ is $(2, 3)$.
Thus,the vertices of the triangle are $A(2, 3), B(1, 0),$ and $C(4, 2)$.