Answer
$$\color {blue}{\bf\text{Yes, the points make a right triangle}}$$
Work Step by Step
For clarity lets start by naming our points:
$A=(-4,3)$
$B=(2,5)$
$C=(-1,-6)$
First, we'll use the distance formula to find the lengths of each side:
$d(P,Q)=\sqrt{(x_{P}-x_{Q})^2+(y_{P}-y_{Q})^2}$
$AB= \sqrt{(-4-2)^2+(3-5)^2}$
$AB= \sqrt{(-6)^2+(-2)^2}$
$AB= \sqrt{36+4}$
$AB= \sqrt{40}$
$AB= 2\sqrt{10}$
$AC= \sqrt{(-4-(-1))^2+(3-(-6))^2}$
$AC= \sqrt{(-3)^2+9^2}$
$AC= \sqrt{9+81}$
$AC= \sqrt{90}$
$AC= 3\sqrt{10}$
$BC= \sqrt{(2-(-1))^2+(5-(-6))^2}$
$BC= \sqrt{3^2+11^2}$
$BC= \sqrt{9+121}$
$BC= \sqrt{130}$
Now that we have the lengths of the sides,
$AB= 2\sqrt{10}$, $AC= 3\sqrt{10}$, $BC= \sqrt{130}$,
we can apply the Pythagorean Theorem $a^2+b^2=c^2$ to see if the sides make a right triangle.
$(2\sqrt{10})^2+(3\sqrt{10})^2= (\sqrt{130})^2$
$4(10)+9(10)= 130$
$40+90= 130$
Which is $\bf \text{true}$, so:
$$\color {blue}{\bf\text{Yes, the points make a right triangle}}$$
