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(B) The centroid $C$ of the triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{

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$(-9, -6)$

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(B) The centroid $C$ of the triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}).$For the given vertices $(3, -1), (1, 3),$ and $(2, 4),$ the centroid $C$ is $(\frac{3+1+2}{3}, \frac{-1+3+4}{3}) = (2, 2).$Next,we find the intersection point $P$ of the lines $x + 3y - 1 = 0$ and $3x - y + 1 = 0.$Solving the system:$x + 3y = 1$$3x - y = -1 \Rightarrow y = 3x + 1$Substituting $y$ into the first equation: $x + 3(3x + 1) = 1$ $\Rightarrow x + 9x + 3 = 1$ $\Rightarrow 10x = -2$ $\Rightarrow x = -\frac{1}{5}.$Then $y = 3(-\frac{1}{5}) + 1 = -\frac{3}{5} + 1 = \frac{2}{5}.$So,$P = (-\frac{1}{5}, \frac{2}{5}).$The line passing through $C(2, 2)$ and $P(-\frac{1}{5}, \frac{2}{5})$ has slope $m = \frac{\frac{2}{5} - 2}{-\frac{1}{5} - 2} = \frac{-\frac{8}{5}}{-\frac{11}{5}} = \frac{8}{11}.$The equation of the line is $y - 2 = \frac{8}{11}(x - 2)$ $\Rightarrow 11y - 22 = 8x - 16$ $\Rightarrow 8x - 11y + 6 = 0.$Checking the options,for $(-9, -6): 8(-9) - 11(-6) + 6 = -72 + 66 + 6 = 0.$Thus,the line passes through $(-9, -6).$

Let $C$ be the centroid of the triangle with vertices $(3, -1), (1, 3),$ and $(2, 4).$ Let $P$ be the point of intersection of the lines $x + 3y - 1 = 0$ and $3x - y + 1 = 0.$ Then the line passing through the points $C$ and $P$ also passes through the point

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