The centroid of a triangle having vertices $\left(x_1,y_1\right), \left(x_2,y_2\right)$ and $\left(x_3,y_3\right)$ is $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right).$

The centroid of the triangle having vertices $(3,-1), (1,3)$ and $(2,4)$ is

$C\left(\frac{3+1+2}{3},\frac{-1+3+4}{3}\right)=(2,2).$

The equation of a line passing through the point on intersection of lines $L_1=0$ and $L_2=0$ is $L_1+\lambda L_2=0.$

Thus, the equation of line which passes through point of intersection $P$ of lines
$x+3 y-1=0$ and $3 x-y+1=0$ is

$(x+3y-1)+\lambda (3x-y+1)=0 ...\left(i\right)$

$\because$ Line $\left(i\right)$ passes through point $C(2,2),$ so

$(2+6-1)+\lambda (6-2+1)=0$

$\Rightarrow 7+5\lambda =0$

$\Rightarrow \lambda =-\frac{7}{5}.$

$\therefore$ Equation of line passes through points $C$ and $P$ is

$(x+3y-1)-\frac{7}{5}(3x-y+1)=0$

$\Rightarrow 5x+15y-5-21x+7y-7=0$

$\Rightarrow -16x+22y-12=0$

$\Rightarrow 8x-11y+6=0.$

From the options, point $(-9,-6)$ satisfy the line.

Hence, option $\left(B\right)$ is correct.