The angle of rotation of the axes so that the equation $x+y-6=0$ may be reduced to the form $X=3\sqrt{2}$ is 

$\theta =?,x+y-6=0\cdots \cdots \cdots \left(1\right)$ (in old coordinates)

It is reduced to the form $X=3\sqrt{2}\cdots \cdots \cdots \left(2\right)$

Substituting $x=X\cos \theta -Y\sin \theta ;y=X\sin \theta +Y\cos \theta$, in eqn(1), we get

$X\cos \theta -Y\sin \theta +X\sin \theta +Y\cos \theta -6=0\cdots \cdots \cdots \left(3\right)$

Equation $\left(2\right)\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}\left(3\right)$ represents same line

$\Rightarrow$Coefficient of Y in equation(3) should be zero

$\Rightarrow -\sin \theta +\cos \theta =0\Rightarrow \cos \theta =\sin \theta \Rightarrow \theta =\frac{\pi }{4}$