Equation of line passing through the point $P(1,2)$ which makes an angle $45^{\circ}$ with the $X$-axis is
$$
\begin{aligned}
(y-2)=m(x-1) \text { and } m & =\tan \theta \\
& =\tan 45^{\circ}=1
\end{aligned}
$$
$\therefore$ Equation is $y-2=1(x-1)$
$$
y-2=x-1 \Rightarrow x-y+1=0
$$
Point of intersection of lines $x-y+1=0$ and $3 x+4 y+5=0$ is $Q$.
Then, for $Q$
$$
x-y+1=0, y=x+1
$$
Put it in $3 x+4 y+5=0$
$$
\begin{aligned}
& \quad 3 x+4(x+1)+5=0 \\
& \Rightarrow \quad 3 x+4 x+4+5=0 \Rightarrow 7 x+9=0 \\
& x=-\frac{9}{7}, y=-\frac{9}{7}+1=-\frac{2}{7} \\
& \therefore Q \text { is }\left(-\frac{9}{7},-\frac{2}{7}\right) \\
& P Q=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \quad \text { [distance formula] } \\
& \quad=\sqrt{\left(1+\frac{9}{7}\right)^2+\left(2+\frac{2}{7}\right)^2}=\sqrt{\left(\frac{16}{7}\right)^2+\left(\frac{16}{7}\right)^2} \\
& \quad=\sqrt{2\left(\frac{16}{7}\right)^2}=\frac{16}{7} \sqrt{2}
\end{aligned}
$$