Let inclination of required line is $\theta$,

So the coordinates of point $B$ can be assumed as

$\left(4-\frac{\sqrt{29}}{3} \cos \theta, 3-\frac{\sqrt{29}}{3} \sin \theta\right)$

Which satisfices $x-y-2=0$

$4-\frac{\sqrt{29}}{3} \cos \theta-3+\frac{\sqrt{29}}{3} \sin \theta-2=0$

$\sin \theta-\cos \theta=\frac{3}{\sqrt{29}}$

By squaring

$\sin 2 \theta=\frac{20}{29}=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$

$\tan \theta=\frac{5}{2}$ only (because slope is greater than 1 )

$$ \sin \theta=\frac{5}{\sqrt{29}}, \cos \theta=\frac{2}{\sqrt{29}} $$

Point $B:\left(\frac{10}{3}, \frac{4}{3}\right)$

Which also satisfies $x+2 y=6$