First, we need to draw the points on the plane to get a better idea of the problem, as shown below
We can obtain the distance between the points and then use the next formula
[tex]\begin{gathered} A=\sqrt[]{s(s-a)(s-b)(s-c)} \\ \text{where} \\ s=\frac{a+b+c}{2} \end{gathered}[/tex]where a, b and c are the sides of the triangle.
In our case,
[tex]\begin{gathered} a=d(AB)=\sqrt[]{(-2-3)^2+(3-6)^2}=\sqrt[]{25+9}=\sqrt[]{34} \\ b=d(BC)=\sqrt[]{(3-2)^2+(6--2)^2}=\sqrt[]{1+64}=\sqrt[]{65} \\ c=d(CA)=\sqrt[]{(2--2)^2+(-2-3)^2}=\sqrt[]{16+25}=\sqrt[]{41} \end{gathered}[/tex]Then, the area is
[tex]\begin{gathered} \Rightarrow s=\frac{\sqrt[]{34}+\sqrt[]{65}+\sqrt[]{41}}{2} \\ \Rightarrow A=\sqrt[]{\frac{\sqrt[]{34}+\sqrt[]{65}+\sqrt[]{41}}{2}(\frac{-\sqrt[]{34}+\sqrt[]{65}+\sqrt[]{41}}{2})(\frac{\sqrt[]{34}-\sqrt[]{65}+\sqrt[]{41}}{2})(\frac{\sqrt[]{34}+\sqrt[]{65}-\sqrt[]{41}}{2})} \end{gathered}[/tex]Simplifying the expression,
[tex]\Rightarrow A=\sqrt[]{(\frac{1}{16})(72+2\sqrt[]{2665})(2\sqrt[]{2665}-72)}[/tex]Therefore,
[tex]\Rightarrow A=\sqrt[]{\frac{1}{16}(4\cdot2665-72^2)}=\sqrt[]{(\frac{1}{16})5476}=\sqrt[]{342.25}[/tex]Finally,
[tex]\Rightarrow A=18.5[/tex]The area of the triangle is 18.5 square units, and the perimeter is
[tex]P=\sqrt[]{34}+\sqrt[]{65}+\sqrt[]{41}=20.296333\ldots\approx20.3[/tex]The perimeter is 20.3 units
