Slope of line $=m$

Slope of $x+y-3=0$ is $-1$

Now, $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\Rightarrow \frac{5}{7}=\left|\frac{m+1}{1-m}\right|$

$\Rightarrow 5\left(1-m\right)=\pm 7\left(m+1\right)$

$\Rightarrow 5-5m=\pm \left(7m+7\right)$

Taking positive sign

$5-5m=7m+7$

$\Rightarrow m=-\frac{1}{6}$

Taking negative sign

$5-5m=-7m-7$

$\Rightarrow m=-6$

Equation of line passing through $\left(1,1\right)$

$\left(I\right)$ $m=-6$

$y-1=-6\left(x-1\right)$

$\Rightarrow 6x+y-7=0$

So, $a=6$, $c=-7$

$\Rightarrow ac=-42$

$\left(II\right)$ $m=-\frac{1}{6}$

$y-1=-\frac{1}{6}\left(x-1\right)$

$\Rightarrow 6y-6=-x+1$

$\Rightarrow \frac{x}{6}+y-\frac{7}{6}=0$

So, $a=\frac{1}{6}$, $c=\frac{-7}{6}$

$\Rightarrow ac=\frac{-7}{36}$