If the angles between the sides of the triang | vedclass

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(B) Let the vertices be $A(2,3,5)$,$B(-1,3,2)$,and $C(3,5,-2)$.Calculate the vectors representing the sides:$\vec{AB} = B - A = (-1-2, 3-3, 2-5) = (-3

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(B) Let the vertices be $A(2,3,5)$,$B(-1,3,2)$,and $C(3,5,-2)$.Calculate the vectors representing the sides:$\vec{AB} = B - A = (-1-2, 3-3, 2-5) = (-3, 0, -3)$.$\vec{BC} = C - B = (3-(-1), 5-3, -2-2) = (4, 2, -4)$.$\vec{CA} = A - C = (2-3, 3-5, 5-(-2)) = (-1, -2, 7)$.Calculate the magnitudes:$|\vec{AB}| = \sqrt{(-3)^2 + 0^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$.$|\vec{BC}| = \sqrt{4^2 + 2^2 + (-4)^2} = \sqrt{16+4+16} = \sqrt{36} = 6$.$|\vec{CA}| = \sqrt{(-1)^2 + (-2)^2 + 7^2} = \sqrt{1+4+49} = \sqrt{54} = 3\sqrt{6}$.Using the Law of Cosines: $\cos \alpha = \frac{|\vec{AB}|^2 + |\vec{AC}|^2 - |\vec{BC}|^2}{2|\vec{AB}||\vec{AC}|}$.Note that $\vec{AC} = -\vec{CA} = (1, 2, -7)$,so $|\vec{AC}| = 3\sqrt{6}$.$\cos \alpha = \frac{18 + 54 - 36}{2(3\sqrt{2})(3\sqrt{6})} = \frac{36}{18\sqrt{12}} = \frac{36}{36\sqrt{3}} = \frac{1}{\sqrt{3}}$.Then $\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \frac{1}{3} = \frac{2}{3}$.Similarly,$\cos \beta = \frac{|\vec{BA}|^2 + |\vec{BC}|^2 - |\vec{AC}|^2}{2|\vec{BA}||\vec{BC}|} = \frac{18 + 36 - 54}{2(3\sqrt{2})(6)} = 0$.So $\beta = 90^\circ$ and $\sin^2 \beta = 1$.Finally,$\sin^2 \gamma = 1 - \sin^2 \alpha - \sin^2 \beta$ is not correct,but $\gamma = 180^\circ - (\alpha + \beta)$.Since $\cos \beta = 0$,$\beta = 90^\circ$,then $\alpha + \gamma = 90^\circ$,so $\gamma = 90^\circ - \alpha$.Thus $\sin^2 \gamma = \sin^2(90^\circ - \alpha) = \cos^2 \alpha = \frac{1}{3}$.Sum $= \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = \frac{2}{3} + 1 + \frac{1}{3} = 2$.

If the angles between the sides of the triangle $ABC$ formed by $A(2,3,5)$,$B(-1,3,2)$ and $C(3,5,-2)$ are $\alpha, \beta$ and $\gamma$,then $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = $

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