The centroid of the triangle formed by the lines $x-3y+3=0,x+3y+3=0,x+y-1=0$ is

Asumming intersection of line $x-3y+3=0 \& x+3y+3=0$ be $A$

Now intersection of line $x+3y+3=0\hspace{0.17em}\& x+y-1=0$ be $B$

And intersection of line $x-3y+3=0 \& x+y-1=0$ be $C$,

Now finding the point $A, B \& C$ by intersecting point we get,

$A\left(-3,0\right), B\left(3,-2\right) \& C\left(0,1\right)$

Now finding the centroid by formula we get $\equiv \left(\frac{-3+3+0}{3},\frac{0-2+1}{3}\right)\equiv \left(0,\frac{-1}{3}\right)$