The triangle formed by the points $A(0,7,10),B(-1,6,6)$ and $C(-4,9,6)$ is:

Use distance formula $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$:$AB=\sqrt{(-1-0{)}^2+(6-7{)}^2+(6-10{)}^2}=\sqrt{1+1+16}=\sqrt{18}$$BC=\sqrt{(-4-(-1){)}^2+(9-6{)}^2+(6-6{)}^2}=\sqrt{9+9+0}=\sqrt{18}$$CA=\sqrt{(0-(-4){)}^2+(7-9{)}^2+(10-6{)}^2}=\sqrt{16+4+16}=\sqrt{36}=6$ Since $AB=BC=\sqrt{18}$, the triangle is Isosceles.Check for right angle using Pythagoras theorem:$A{B}^2+B{C}^2=18+18=36$$C{A}^2=6^2=36$ Since $A{B}^2+B{C}^2=C{A}^2$, it is a right-angled triangle.Therefore, the triangle is a right-angled isosceles triangle.