Verify that the points $(0, 7, -10)$,$(1, 6, -6)$,and $(4, 9, -6)$ are the vertices of an isosceles triangle.

Let the points be $A(0, 7, -10)$,$B(1, 6, -6)$,and $C(4, 9, -6)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$:
$AB = \sqrt{(1 - 0)^2 + (6 - 7)^2 + (-6 - (-10))^2} = \sqrt{1^2 + (-1)^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$.
$BC = \sqrt{(4 - 1)^2 + (9 - 6)^2 + (-6 - (-6))^2} = \sqrt{3^2 + 3^2 + 0^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$.
$CA = \sqrt{(0 - 4)^2 + (7 - 9)^2 + (-10 - (-6))^2} = \sqrt{(-4)^2 + (-2)^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$.
Since $AB = BC = 3\sqrt{2}$ and $AB \neq CA$,two sides of the triangle are equal.
Therefore,the given points are the vertices of an isosceles triangle.