Answer
$${\rm{7 unit}}{{\rm{s}}^2}$$
Work Step by Step
$$\eqalign{
& {\rm{Let\, the\, points\, }}P\left( {\underbrace 2_{{x_1}},\underbrace { - 2}_{{y_1}}} \right),\,\,Q\left( {\underbrace 0_{{x_2}},\underbrace 0_{{y_2}}} \right),\,\,R\left( {\underbrace { - 3}_{{x_3}},\underbrace { - 4}_{{y_3}}} \right) \cr
& {\rm{Calcule\, the\, area \,using}} \cr
& D = {1 \over 2}\left| {\matrix{
{{x_1}} & {{y_1}} & 1 \cr
{{x_2}} & {{y_2}} & 1 \cr
{{x_3}} & {{y_3}} & 1 \cr
} } \right| \cr
& {\rm{Substitute\, the\, values}} \cr
& 2D = \left| {\matrix{
2 & { - 2} & 1 \cr
0 & 0 & 1 \cr
{ - 3} & { - 4} & 1 \cr
} } \right| \cr
& {\rm{Calculate\, the\, determinant }} \cr
& 2D = 0\left| {\matrix{
{ - 2} & 1 \cr
{ - 4} & 1 \cr
} } \right| + 0\left| {\matrix{
2 & 1 \cr
{ - 3} & 1 \cr
} } \right| + 1\left| {\matrix{
2 & { - 2} \cr
{ - 3} & { - 4} \cr
} } \right| \cr
& 2D = 0 + 0 + \left( { - 8 - 6} \right) \cr
& 2D = - 14 \cr
& D = - 7 \cr
& {\rm{Calculate\, the\, absolute\, value\, of\, }}D \cr
& D = 7 \cr
& {\rm{The\, area\, of \,the\, triangle\, is \,7 \,unit}}{{\rm{s}}^2} \cr} $$
