Consider a triangle $$ABC$$ having the vertices $$A(1, 2)$$, $$B(\alpha, \beta)$$ and $$C(\gamma, \delta)$$ and angles $$\angle ABC = \frac{\pi}{6}$$ 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 _____

We are given a triangle $$ABC$$ with $$A(1, 2)$$, $$B(\alpha, \beta)$$, $$C(\gamma, \delta)$$, where $$B$$ and $$C$$ lie on the line $$y = x + 4$$. We know $$\angle ABC = \frac{\pi}{6}$$ and $$\angle BAC = \frac{2\pi}{3}$$.

From the sum of angles in a triangle we have $$\angle BCA = \pi - \frac{2\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$$. Since $$\angle ABC = \angle BCA = \frac{\pi}{6}$$, the triangle is isosceles with $$AB = AC$$.

The line $$y = x + 4$$ can be written in standard form as $$x - y + 4 = 0$$. Therefore the perpendicular distance from $$A(1, 2)$$ to this line is given by $$d = \frac{|1 - 2 + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{3}{\sqrt{2}}$$.

Because the triangle is isosceles, the perpendicular from $$A$$ to $$BC$$ also bisects $$BC$$. Hence the foot of the perpendicular, $$M$$, from $$(1, 2)$$ to the line $$x - y + 4 = 0$$ is calculated by projecting onto the normal direction, yielding $$M = \left(1 - \frac{1 \cdot (1 - 2 + 4)}{2},\; 2 - \frac{(-1)(1 - 2 + 4)}{2}\right) = \left(-\tfrac{1}{2},\; \tfrac{7}{2}\right)$$.

In right triangle $$ABM$$, the angle at $$B$$ is $$\angle ABM = \frac{\pi}{6}$$ and the length $$AM$$ equals the perpendicular distance $$\frac{3}{\sqrt{2}}$$. Using the tangent function, $$\tan\bigl(\tfrac{\pi}{6}\bigr) = \frac{AM}{BM} \implies \frac{1}{\sqrt{3}} = \frac{3/\sqrt{2}}{BM}$$, which gives $$BM = \frac{3\sqrt{3}}{\sqrt{2}}$$.

The direction vector along the line $$y = x + 4$$ is $$\tfrac{1}{\sqrt{2}}(1, 1)$$. Since $$B$$ and $$C$$ lie on opposite sides of $$M$$ at distance $$BM$$ along this line, their coordinates are

$$B = M + \frac{3\sqrt{3}}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}(1, 1) = \left(-\tfrac{1}{2} + \tfrac{3\sqrt{3}}{2},\; \tfrac{7}{2} + \tfrac{3\sqrt{3}}{2}\right),$$

$$C = M - \frac{3\sqrt{3}}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}(1, 1) = \left(-\tfrac{1}{2} - \tfrac{3\sqrt{3}}{2},\; \tfrac{7}{2} - \tfrac{3\sqrt{3}}{2}\right).$$

Thus $$\alpha = -\tfrac{1}{2} + \tfrac{3\sqrt{3}}{2}$$ and $$\gamma = -\tfrac{1}{2} - \tfrac{3\sqrt{3}}{2}$$. To compute $$\alpha^2 + \gamma^2$$ we use the identity $$(a+b)^2 + (a-b)^2 = 2(a^2 + b^2)$$ with $$a = -\tfrac{1}{2}$$ and $$b = \tfrac{3\sqrt{3}}{2}$$, giving

$$\alpha^2 + \gamma^2 = \left(-\tfrac{1}{2} + \tfrac{3\sqrt{3}}{2}\right)^2 + \left(-\tfrac{1}{2} - \tfrac{3\sqrt{3}}{2}\right)^2 = 2\left(\tfrac{1}{4} + \tfrac{27}{4}\right) = 14.$$

The answer is $$\boxed{14}$$.