Given equations are
$x+y=4 ...(i)$
and $\quad x-y=2$ ...(ii)
From Eqs. (i) and $(ii),$ we get
$x=3$ and $y=1$
The line through this point making an angle tan $^{-1} \frac{3}{4}$ with the $X$ -axis is
$(y-1)=\frac{3}{4}(x-3) \quad\left[\because m=\frac{3}{4}\right]$
$\Rightarrow \quad y=\frac{3 x}{4}-\frac{5}{4}=\frac{3 x-5}{4}$
Since, this line intersects the parabola
$y^{2}=4(x-3)$ at points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, respectively.
$\therefore$ Putting $y=\frac{3 x-5}{4}$ in equation of parabola, we get
$$
\left(\frac{3 x-5}{4}\right)^{2}=4(x-3)
$$
$\Rightarrow \quad 9 x^{2}-94 x+217=0$
$\Rightarrow \quad x_{1}+x_{2}=\frac{94}{9}$ and $x_{1} x_{2}=\frac{217}{9}$
$\begin{aligned} \therefore \quad\left|x_{1}-x_{2}\right| &=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\ &=\sqrt{\left(\frac{94}{9}\right)^{2}-4 \times \frac{217}{9}} \\ &=\frac{32}{9} \end{aligned}$