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(D) The line $BC$ has the equation $y = x + 4$,so its slope is $m_1 = 1$. Let the slopes of lines $AB$ and $AC$ be $m_2$ and $m_3$ respectively. The a

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$14$

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(D) The line $BC$ has the equation $y = x + 4$,so its slope is $m_1 = 1$. Let the slopes of lines $AB$ and $AC$ be $m_2$ and $m_3$ respectively. The angle between $AB$ and $BC$ is $\frac{\pi}{6}$,so $	an(\frac{\pi}{6}) = |\frac{m_2 - 1}{1 + m_2}| = \frac{1}{\sqrt{3}}$. This gives $m_2 = 2 \pm \sqrt{3}$.The line $AB$ passes through $A(1, 2)$ with slope $m_2$. Its equation is $y - 2 = m_2(x - 1)$.For $m_2 = 2 + \sqrt{3}$,$y - 2 = (2 + \sqrt{3})(x - 1)$. Solving with $y = x + 4$ gives $x + 2 = (2 + \sqrt{3})x - 2 - \sqrt{3}$,so $x = \frac{4 + \sqrt{3}}{1 + \sqrt{3}} = \frac{(4 + \sqrt{3})(\sqrt{3} - 1)}{2} = \frac{4\sqrt{3} - 4 + 3 - \sqrt{3}}{2} = \frac{3\sqrt{3} - 1}{2}$.For $m_2 = 2 - \sqrt{3}$,$y - 2 = (2 - \sqrt{3})(x - 1)$. Solving with $y = x + 4$ gives $x + 2 = (2 - \sqrt{3})x - 2 + \sqrt{3}$,so $x = \frac{4 - \sqrt{3}}{1 - \sqrt{3}} = \frac{(4 - \sqrt{3})(1 + \sqrt{3})}{-2} = \frac{4 + 4\sqrt{3} - \sqrt{3} - 3}{-2} = \frac{1 + 3\sqrt{3}}{-2}$.Thus,$\alpha$ and $\gamma$ are $\frac{3\sqrt{3} - 1}{2}$ and $\frac{-(3\sqrt{3} + 1)}{2}$.$\alpha^2 + \gamma^2 = \frac{27 + 1 - 6\sqrt{3}}{4} + \frac{27 + 1 + 6\sqrt{3}}{4} = \frac{28 + 28}{4} = 14$.

Consider a triangle $ABC$ having the vertices $A(1,2)$,$B(\alpha, \beta)$,and $C(\gamma, \delta)$. The angles are $\angle ABC = \frac{\pi}{6}$ and $\angle BAC = \frac{2\pi}{3}$. If the points $B$ and $C$ lie on the line $y = x + 4$,then $\alpha^2 + \gamma^2$ is equal to:

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