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(D) Let the sides be $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = 2\sqrt{3}\hat{i} - 2\sqrt{3}\hat{j} + \sqrt{3}\hat{k}$.Magnitude $|\vec{

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$\frac{\pi}{6} ; 9+3\sqrt{3}$

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(D) Let the sides be $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = 2\sqrt{3}\hat{i} - 2\sqrt{3}\hat{j} + \sqrt{3}\hat{k}$.Magnitude $|\vec{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4+1+4} = 3$.Magnitude $|\vec{b}| = \sqrt{(2\sqrt{3})^2 + (-2\sqrt{3})^2 + (\sqrt{3})^2} = \sqrt{12+12+3} = \sqrt{27} = 3\sqrt{3}$.The third side is $\vec{c} = \vec{b} - \vec{a} = (2\sqrt{3}-2)\hat{i} + (-2\sqrt{3}-1)\hat{j} + (\sqrt{3}+2)\hat{k}$.Magnitude $|\vec{c}|^2 = (2\sqrt{3}-2)^2 + (-2\sqrt{3}-1)^2 + (\sqrt{3}+2)^2 = (12+4-8\sqrt{3}) + (12+1+4\sqrt{3}) + (3+4+4\sqrt{3}) = 16+13+7 = 36$.So, $|\vec{c}| = 6$.The sides are $3, 3\sqrt{3}, 6$.Perimeter $= 3 + 3\sqrt{3} + 6 = 9 + 3\sqrt{3}$.Using the Law of Cosines, the angles are $\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{27+36-9}{2(3\sqrt{3})(6)} = \frac{54}{36\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \implies A = 30^\circ = \frac{\pi}{6}$.Since $3 < 3\sqrt{3} < 6$, the least angle is opposite the smallest side $3$, which is $\frac{\pi}{6}$.

Two adjacent sides of a triangle are represented by the vectors $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = 2\sqrt{3}\hat{i} - 2\sqrt{3}\hat{j} + \sqrt{3}\hat{k}$. Then the least angle of the triangle and the perimeter of the triangle are respectively:

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