In an isosceles triangle $$ABC$$, the vertex $$A$$ is $$(6, 1)$$ and the equation of the base $$BC$$ is $$2x + y = 4$$. Let the point $$B$$ lie on the line $$x + 3y = 7$$. If $$(\alpha, \beta)$$ is the centroid of $$\triangle ABC$$, then $$15(\alpha + \beta)$$ is equal to

We are given an isosceles triangle $$ABC$$ with vertex $$A = (6, 1)$$, base $$BC$$ on the line $$2x + y = 4$$, and point $$B$$ on the line $$x + 3y = 7$$.

Since point $$B$$ lies on both lines, from $$x + 3y = 7$$ we have $$x = 7 - 3y$$, and substituting into $$2x + y = 4$$ gives $$2(7 - 3y) + y = 4$$, so $$14 - 6y + y = 4$$, $$-5y = -10$$, $$y = 2, \quad x = 7 - 6 = 1$$. Thus, $$B = (1, 2)$$.

Because the triangle is isosceles with vertex $$A$$, we require $$AB = AC$$. We compute $$AB^2 = (6-1)^2 + (1-2)^2 = 25 + 1 = 26$$. Letting $$C = (c, 4 - 2c)$$ so that it lies on $$2x + y = 4$$, we have $$AC^2 = (6-c)^2 + (1-(4-2c))^2 = (6-c)^2 + (2c-3)^2 = 36 - 12c + c^2 + 4c^2 - 12c + 9 = 5c^2 - 24c + 45$$. Setting this equal to 26 leads to $$5c^2 - 24c + 45 = 26$$, or $$5c^2 - 24c + 19 = 0$$. By the quadratic formula, $$c = \frac{24 \pm \sqrt{576 - 380}}{10} = \frac{24 \pm \sqrt{196}}{10} = \frac{24 \pm 14}{10}$$, giving $$c = \frac{38}{10} = \frac{19}{5} \quad \text{or} \quad c = \frac{10}{10} = 1$$. The case $$c = 1$$ corresponds to point $$B$$, so we take $$c = \frac{19}{5}$$, whence $$C = \left(\frac{19}{5}, 4 - \frac{38}{5}\right) = \left(\frac{19}{5}, -\frac{18}{5}\right)$$.

The centroid $$(\alpha, \beta)$$ of triangle $$ABC$$ is given by the averages of the coordinates, namely $$\alpha = \frac{6 + 1 + \frac{19}{5}}{3} = \frac{\frac{30 + 5 + 19}{5}}{3} = \frac{54}{15} = \frac{18}{5}$$ and $$\beta = \frac{1 + 2 + \left(-\frac{18}{5}\right)}{3} = \frac{\frac{5 + 10 - 18}{5}}{3} = \frac{-3}{15} = -\frac{1}{5}$$.

Finally, since $$\alpha + \beta = \frac{18}{5} - \frac{1}{5} = \frac{17}{5}$$, we have $$15(\alpha + \beta) = 15 \times \frac{17}{5} = 3 \times 17 = 51$$. The correct answer is Option A: $$51$$.