Answer
b) The square angle is at $A(2,1,6)$
c) $\operatorname{Area}=49(\text { units })^{2}$
a) $A B C$ is a square triangle, we have proven that
Work Step by Step
a) Let us consider the points
\[
C(8,5,-6), B(4,7,9) \text { and } A(2,1,6)
\]
We compute the three distances:
\[
\begin{array}{l}
d(A, B)=\overline{A B}=\sqrt{(-1+7)^{2}+(-6+9)^{2}+(-2+4)^{2}}=\sqrt{9+36+4}=\sqrt{49}=7 \\
d(A, C)=\overline{A C}=\sqrt{(-2+8)^{2}+(-1+5)^{2}+(-6-6)^{2}}=\sqrt{144+16+36}=14 \\
d(B, C)=\overline{B C}=\sqrt{(-4+8)^{2}+(-7+5)^{2}+(-9-6)^{2}}=\sqrt{225+4+16}=7 \sqrt{5}
\end{array}
\]
b) $\operatorname{Since} \overline{B C}^{2}=(7 \sqrt{5})^{2}=14^{2}+7^{2}=A C^{2}, A B C+ A B^{2} $ is a right triangle, which has the right angle at $A(2,1,6)$
c) Since $A B C$ is a right triangle and the hypotenuse is $\overline{B C},$ the area is:
\[
\text { Area }=\frac{1}{2} \overline{A B} \cdot \overline{A C}=\frac{1}{2}(14)(7)=49(\text { units })^{2}
\]
