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Q.
The points $(0,7,10),(-1,6,6)$ and $(-4,9,6)$ form
Introduction to Three Dimensional Geometry
Solution:
Let $P (0,7,10),\, Q (-1,6,6)$ and $R (-4,9,6)$ be the vertices of a triangle
Here, $P Q=\sqrt{1+1+16}=3 \sqrt{2}$
$Q R=\sqrt{9+9+0}=3 \sqrt{2}$
$P R= \sqrt{16+4+16}=6$
Now, $P Q^{2}+Q R^{2}=(3 \sqrt{2})^{2}+(3 \sqrt{2})^{2}=36=( PR )^{2}$
Therefore, $\Delta PQR$ is a right angled triangle at $Q$. Also, $OQ$ $=$ OR. Hence, $\Delta PQR$ is a right angled isosceles triangle.