If $(a, b)$ is the centre of the circle passing through the vertices of the triangle formed by $x+y=6, 2x+y=4$ and $x+2y=5,$ then $\left(a, b\right)$ is

$x+y=6......\left(1\right), 2x+y=4.......\left(2\right),x+2y=5......\left(3\right)$

First apply elimination method on $\left(1\right) \& \left(2\right)$ we get,

$x=-2,y=8$

Now take $\left(2\right) \& \left(3\right)$,we get

$x=1,y=2$

And then take $\left(1\right) \& \left(3\right)$ we get

$x=7,y=-1$

The vertices of triangle are the point of intersection of given three lines $\left(-2,8\right),\left(1,2\right),\left(7,-1\right)$

Let the equation of circle be $x^2+y^2+2gx+2fy+c=0$

We are given that the circle passes through vertices then put the found vertices value in equation of circle we get

$16f-4g+c+68=0 .....\left(4\right)$

$2g+4f+c+5=0....\left(5\right) or c=-2g-4f-5$

$14g-2f+c+50=0.....\left(6\right)$

Now put $c$ found from $\left(5\right)$ in $\left(4\right) \& \left(6\right)$ we get

$12f-6g+63=0......\left(7\right)$

$12g-6f+45=0........\left(8\right)$

Apply elimination method,we get

$g=\frac{-17}{2},f=-\frac{19}{2}$

So centre is $\left(\frac{17}{2},\frac{19}{2}\right)$