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(B) The vertices are $A(2, 2, 1)$,$B(1, 2, 2)$,and $C(2, 1, 2)$.Calculate the side lengths:$AB = \sqrt{(1-2)^2 + (2-2)^2 + (2-1)^2} = \sqrt{1+0+1} = \

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$\frac{1}{2}$

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(B) The vertices are $A(2, 2, 1)$,$B(1, 2, 2)$,and $C(2, 1, 2)$.Calculate the side lengths:$AB = \sqrt{(1-2)^2 + (2-2)^2 + (2-1)^2} = \sqrt{1+0+1} = \sqrt{2}$$BC = \sqrt{(2-1)^2 + (1-2)^2 + (2-2)^2} = \sqrt{1+1+0} = \sqrt{2}$$CA = \sqrt{(2-2)^2 + (2-1)^2 + (1-2)^2} = \sqrt{0+1+1} = \sqrt{2}$Since $AB = BC = CA = \sqrt{2}$,the triangle is equilateral.For an equilateral triangle,the orthocenter $H$ coincides with the centroid $G$.$G = \left(\frac{2+1+2}{3}, \frac{2+2+1}{3}, \frac{1+2+2}{3}ight) = \left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}ight)$.In an equilateral triangle,the length of the perpendicular from the centroid to any side is the distance from the centroid to the midpoint of that side.For side $AB$,the midpoint $D$ is $\left(\frac{2+1}{2}, \frac{2+2}{2}, \frac{1+2}{2}ight) = \left(\frac{3}{2}, 2, \frac{3}{2}ight)$.$l_1 = 	ext{distance } GD = \sqrt{\left(\frac{5}{3}-\frac{3}{2}ight)^2 + \left(\frac{5}{3}-2ight)^2 + \left(\frac{5}{3}-\frac{3}{2}ight)^2}$$l_1 = \sqrt{\left(\frac{1}{6}ight)^2 + \left(-\frac{1}{3}ight)^2 + \left(\frac{1}{6}ight)^2} = \sqrt{\frac{1}{36} + \frac{4}{36} + \frac{1}{36}} = \sqrt{\frac{6}{36}} = \sqrt{\frac{1}{6}}$.Since the triangle is equilateral,$l_1 = l_2 = l_3 = \sqrt{\frac{1}{6}}$.Therefore,$l_1^2 + l_2^2 + l_3^2 = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$.

Similar Questions

Statement $(A):$ If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = 0$,then $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$.
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