If $\alpha ,\beta$are the roots of the quadratic equation $x^2-5x+6=0$ such that $\alpha <\beta$ and the vertices of a triangle are $A\left(3,5,-2\right),B\left(\alpha ,\beta ,5\right),C\left(-1,\beta ,\alpha \right)$,then the triangle is

The roots of the equation $x^2-5x+6=0$ are $\alpha =2,\beta =3$

The vertices of triangle are$A\left(3,5,-2\right),B\left(2,3,5\right),C\left(-1,3,2\right)$

The distance between two points $A\left(x_1,y_1,z_1\right)$ and $B\left(x_2,y_2,z_2\right)$ is$AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

Hence, the sides of the triangle are 

$\begin{array}{c}AB=\sqrt{1+4+49}=\sqrt{54}\\ BC=\sqrt{9+0+9}=\sqrt{18}\\ CA=\sqrt{16+4+16}=\sqrt{36}\end{array}$

Observing the lengths of sides of the triangle one is square of the sum of the squares of  the other two

Hence the triangle is right angle triangle.