(1)

$|\overrightarrow{AB}| = \sqrt{4+4+4} = \sqrt{12} \qquad |\overrightarrow{BC}| = \sqrt{36+1+4} = \sqrt{41} \qquad |\overrightarrow{CA}| = \sqrt{16+9+16} = \sqrt{41}$

$\cos\theta_{\angle CAB} = \frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|} = \frac{(2,2,-2)\cdot(-4,3,-4)}{\sqrt{12}\sqrt{41}} = \frac{-8+6+8}{\sqrt{12}\sqrt{41}} = \frac{6}{\sqrt{12}\sqrt{41}}$

$\cos\theta_{\angle ABC} = \frac{\overrightarrow{BA}\cdot\overrightarrow{BC}}{|\overrightarrow{BA}||\overrightarrow{BC}|} = \frac{(-2,-2,2)\cdot(-6,1,-2)}{\sqrt{12}\sqrt{41}} = \frac{12-2-4}{\sqrt{12}\sqrt{41}} = \frac{6}{\sqrt{12}\sqrt{41}}$

$\cos\theta_{\angle ACB} = \frac{\overrightarrow{CB}\cdot\overrightarrow{CA}}{|\overrightarrow{CB}||\overrightarrow{CA}|} = \frac{(4,-3,4)\cdot(6,-1,2)}{\sqrt{41}\sqrt{41}} = \frac{24+3+8}{41} = \frac{35}{41}$

これから $|\overrightarrow{BC}| = |\overrightarrow{CA}|$ の二等辺三角形

(2)

$|\overrightarrow{AB}| = \sqrt{16+0+25} = \sqrt{41} \qquad |\overrightarrow{BC}| = \sqrt{81+1+1} = \sqrt{83} \qquad |\overrightarrow{CA}| = \sqrt{25+1+16} = \sqrt{42}$

$\cos\theta_{\angle CAB} = \frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|} = \frac{(4,0,-5)\cdot(-5,1,-4)}{\sqrt{41}\sqrt{42}} = \frac{-20+20}{\sqrt{41}\sqrt{42}} = 0$

$\cos\theta_{\angle ABC} = \frac{\overrightarrow{BA}\cdot\overrightarrow{BC}}{|\overrightarrow{BA}||\overrightarrow{BC}|} = \frac{(-4,0,5)\cdot(-9,1,1)}{\sqrt{41}\sqrt{83}} = \frac{36+5}{\sqrt{41}\sqrt{83}} = \frac{41}{\sqrt{41}\sqrt{83}}$

$\cos\theta_{\angle ACB} = \frac{\overrightarrow{CB}\cdot\overrightarrow{CA}}{|\overrightarrow{CB}||\overrightarrow{CA}|} = \frac{(5,-1,4)\cdot(9,-1,-1)}{\sqrt{42}\sqrt{83}} = \frac{45+1-4}{\sqrt{42}\sqrt{83}} = \frac{42}{\sqrt{42}\sqrt{83}}$

これから∠A＝90° の直角三角形

($3$)

$\left|\overrightarrow{AB}\right| = \sqrt{(a-b)^2 + (b-c)^2 + (c-a)^2} \qquad \left|\overrightarrow{BC}\right| = \sqrt{(b-c)^2 + (c-a)^2 + (a-b)^2}$

$\left|\overrightarrow{CA}\right| = \sqrt{(c-a)^2 + (a-b)^2 + (b-c)^2}$

これから $\left|\overrightarrow{AB}\right| = \left|\overrightarrow{BC}\right| = \left|\overrightarrow{CA}\right|$ の正三角形