To find $B$

$3 x-2 y=-1$ and $x+y=3$

solving above two equations we get $y=2$ and $x=1$

$B=(1,2)$

To find $C$

$3 x-2 y=-1$ and $x-3 y=-9$

solving above two equations we get $y=27 / 6$ and $x=15 / 7$

$C=\left(\frac{15}{7}, \frac{27}{6}\right)$

$s=\frac{\text {perimeteroftriangle}}{2}$

$a=A B=\sqrt{(1-0)^{2}+(2-3)^{2}}$

$=\sqrt{1+1}$

$=\sqrt{2}$

$b=B C=\sqrt{\left(\frac{15}{7}-1\right)^{2}+\left(\frac{26}{7}-2\right)^{2}}$

$=\sqrt{\left(\frac{8}{7}\right)^{2}+\left(\frac{12}{7}\right)^{2}}$

$\frac{4 \sqrt{13}}{7}$

$c=C A=\sqrt{\left(\frac{15}{7}-0\right)^{2}+\left(\frac{26}{7}-3\right)^{2}}$

$=\sqrt{\left(\frac{15}{7}\right)^{2}+\left(\frac{5}{7}\right)^{2}}$

$=\frac{5 \sqrt{10}}{7}$

$s=\frac{a+b+c}{2}$

$s=\frac{\sqrt{2}+\frac{4 \sqrt{13}}{7}+\frac{5 \sqrt{10}}{7}}{2}$

$s=\frac{14 \sqrt{2}+4 \sqrt{13}+5 \sqrt{10}}{14}$

$s-a=\frac{-7 \sqrt{2}+4 \sqrt{13}+5 \sqrt{10}}{14}$

$s-b=\frac{14 \sqrt{2}-4 \sqrt{13}+5 \sqrt{10}}{14}$

$s-c=\frac{14 \sqrt{2}+4 \sqrt{13}-5 \sqrt{10}}{14}$

$A=\sqrt{s(s-a)(s-b)(s-c)}$

$A=\sqrt{\left(\frac{-7 \sqrt{2}+4 \sqrt{13}+5 \sqrt{10}}{14}\right)\left(\frac{14 \sqrt{2}-4 \sqrt{13}+5 \sqrt{10}}{14}\right)\left(\frac{14 \sqrt{2}+4 \sqrt{13}-5 \sqrt{10}}{14}\right)}$

Multiplying and solving

$A=\frac{16}{7}$ sq unit