Given equations of pair of straight lines, $x y-x-y+1=0$ can be rewritten as
$\begin{aligned} & x(y-1)-1(y-1)=0 \\ & \Rightarrow \quad(x-1)(y-1)=0 \\ & \Rightarrow \quad x=1 \text { and } y=1 \\ & \end{aligned}$
Thus, the three concurrent lines are
$$
\begin{aligned}
& x=1 \\
& y=1
\end{aligned}
$$
and
$$
x+a y-3=0
$$
Since, Eqs. (i) and (ii), intersect at only point, namely $(1,1)$, therefore this point also satisfy the Eq. (iii), we get
$$
\Rightarrow \quad \begin{aligned}
1+a-3 & =0 \\
a & =2
\end{aligned}
$$
Now, the pair of lines
$$
\begin{aligned}
& a x^2-13 x y-7 y^2+x+23 y-6=0 \text { becomes } \\
& 2 x^2-13 x y-7 y^2+x+23 y-6=0
\end{aligned}
$$

The acute angle $\theta$ between these lines is given by
$$
\begin{aligned}
& \tan \theta=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|=\left|\frac{2 \sqrt{\left(-\frac{13}{2}\right)^2+14}}{-5}\right| \\
& =\left|\frac{2 \sqrt{\frac{169+56}{4}}}{-5}\right|=\left|\frac{2 \sqrt{\frac{225}{4}}}{-5}\right|=\left|\frac{2 \times \frac{15}{2}}{-5}\right|=|-3|=3 \\
& \theta=\tan ^{-1}(3)=\cos ^{-1}\left(\frac{1}{\sqrt{1+3^2}}\right)=\cos ^{-1}\left(\frac{1}{\sqrt{10}}\right)
\end{aligned}
$$