The point-slope form of linear equation is:
[tex]y-y_1=m(x-x_1)[/tex]
Hence,
[tex]\displaystyle y-5= \frac{2}{3}(x+2)\\\\y= \frac{2}{3}x+5 \frac{4}{3} \\\\y=\frac{2}{3}x+6 \frac{1}{3} [/tex]
The points A,B are lying on the line, if and only if the next equality is true:
[tex]\displaystyle y=\frac{2}{3}x+6 \frac{1}{3} [/tex]
Therefore,
[tex]\displaystyle A(x,3) \Rightarrow\\\\3=\frac{2}{3}x+6 \frac{1}{3} \\\\9=2x+19\\\\-10=2x\\\\x=-5[/tex]
[tex]\displaystyle B(-2,y)\Rightarrow \\\\y=\frac{-4}{3}+6 \frac{1}{3} \\\\y=5[/tex]
