The mid-points of the sides of a triangle are $D\left(6,1\right),E\left(3,5\right)$ and $F\left(-1,-2\right)$ then the vertex opposite to $D$ is

Given, midpoints of sides of a triangle are $D\left(6,1\right),E\left(3,5\right),F\left(-1,-2\right)$

Let $ABC$  is a triangle $A=\left(x_1,y_1\right)$

$D,E,F$  are midpoints of $BC,AC\text{and}AB$

By midpoint theorem, $DE=AF,DE\left|\right|AF$

$DE=AF,DF\left|\right|AE$

$\therefore AEDF$  is a parallelogram

$\therefore$  Midpoint of $AD=$  Midpoint of $EF$

$\left(\frac{x_1+6}{2},\frac{y_1+1}{2}\right)=\left(\frac{3-1}{2},\frac{5-2}{2}\right)$

By comparison

$x_1=3-1-6$              $y_1=5-2-1$

$x_1=-4$                      $y_1=2$

$A=\left(x_1,y_1\right)=\left(-4,2\right)$