Given point $B$ lie on the line $x-y-2=0$, so the point will be $B\left(x_1,x_1-2\right)$

Now given distance between $AB$ is  $\frac{\sqrt{29}}{3}$, so by distance formula,

$\sqrt{{\left(x_1-4\right)}^2+{\left(x_1-2-3\right)}^2}=\frac{\sqrt{29}}{3}$

Now on squaring on both side we get,

$18x_1^2-162x_1+340=0$

On solving we get, $x_1=\frac{51}{9}$ or $x_1=\frac{10}{3}$

So, $y_1=\frac{33}{9}$ or $y_1=\frac{4}{3}$

Now we can see only $x+2y=6$ will satisfy the point $\left(\frac{10}{3},\frac{4}{3}\right)$