1. Understand the geometry

In an isosceles triangle with vertex at $A$, the base is $BC$. Hence the equal sides are: $AB = AC$ So the perpendicular from $A$ to the base $BC$ will pass through the midpoint of $BC$.

Thus, if $M$ is the midpoint of $BC$, then $AM \perp BC$.

2. Equation of the base $BC$

Given: $2x + y = 4$ Its slope is $-2$.

Therefore, any line perpendicular to it has slope: $\frac{1}{2}$

The perpendicular from $A(6,1)$ to $BC$ is: $y - 1 = \frac{1}{2}(x - 6)$ $2y - 2 = x - 6$ $x - 2y - 4 = 0$

So the median/altitude from $A$ is: $x - 2y = 4$

3. Find midpoint $M$ of $BC$

Since $M$ lies on both:

• base $BC: 2x + y = 4$
• altitude/median: $x - 2y = 4$

Solve: $2x + y = 4 \quad ...(1)$ $x - 2y = 4 \quad ...(2)$

From (2): $x = 4 + 2y$ Substitute into (1): $2(4+2y)+y=4$ $8+5y=4$ $5y=-4$ $y=-\frac{4}{5}$ Then $x=4+2\left(-\frac45\right)=4-\frac85=\frac{12}{5}$

Thus, $M\left(\frac{12}{5}, -\frac{4}{5}\right)$

4. Find point $B$

Point $B$ lies on both:

• base $BC: 2x+y=4$
• line $x+3y=7$

Solve: $2x+y=4 \quad ...(1)$ $x+3y=7 \quad ...(2)$

From (1): $y=4-2x$ Substitute into (2): $x+3(4-2x)=7$ $x+12-6x=7$ $-5x=-5$ $x=1$ Then $y=4-2(1)=2$

So, $B=(1,2)$

5. Use midpoint to find $C$

If $M$ is the midpoint of $BC$, then $M=\left(\frac{x_B+x_C}{2},\frac{y_B+y_C}{2}\right)$

So, $x_C = 2x_M - x_B = 2\cdot \frac{12}{5} - 1 = \frac{24}{5}-\frac{5}{5}=\frac{19}{5}$ $y_C = 2y_M - y_B = 2\cdot \left(-\frac45\right)-2 = -\frac85-\frac{10}{5}=-\frac{18}{5}$

Hence, $C=\left(\frac{19}{5}, -\frac{18}{5}\right)$

6. Find the centroid $(\alpha,\beta)$

Centroid of triangle with vertices $A(6,1)$, $B(1,2)$, $C\left(\frac{19}{5},-\frac{18}{5}\right)$ is: $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$

So, $\alpha = \frac{6+1+\frac{19}{5}}{3} = \frac{\frac{30}{5}+\frac{5}{5}+\frac{19}{5}}{3} = \frac{54/5}{3} = \frac{18}{5}$

$\beta = \frac{1+2-\frac{18}{5}}{3} = \frac{3-\frac{18}{5}}{3} = \frac{\frac{15}{5}-\frac{18}{5}}{3} = \frac{-3/5}{3} = -\frac{1}{5}$

Thus, $\alpha+\beta = \frac{18}{5} - \frac{1}{5} = \frac{17}{5}$

Therefore, $15(\alpha+\beta)=15\cdot \frac{17}{5}=3\cdot 17=51$

7. Check options

The value is: $\boxed{51}$ So the correct option is: $\boxed{\text{C}}$