3. Let line $l$ pass through point $M(1,2)$. If line $l$ is intercepted by the two parallel lines $4 x+3 y+1=0$ and $4 x+3 y+6=0$ to form a segment of length $\sqrt{2}$, then the equation of line $l$ is $\qquad$
3. $x+7 y=15$ or $7 x-y=5$.

Let the line $l: y-2=k(x-1)$.
Solve for the intersection points by combining this equation with the two parallel line equations in the problem:
$$
\begin{array}{l}
A\left(\frac{3 k-7}{3 k+4}, \frac{-5 k+8}{3 k+4}\right), \\
B\left(\frac{3 k-12}{3 k+4}, \frac{-10 k+8}{3 k+4}\right) .
\end{array}
$$

Given $|A B|=\sqrt{2}$
$$
\begin{array}{l}
\Rightarrow \sqrt{\left(\frac{5}{3 k+4}\right)^{2}+\left(\frac{5 k}{3 k+4}\right)^{2}}=\sqrt{2} \\
\Rightarrow \frac{25\left(k^{2}+1\right)}{(3 k+4)^{2}}=2 \\
\Rightarrow k_{1}=7, k_{2}=-\frac{1}{7} .
\end{array}
$$

Substituting back into the assumed equation yields $x+7 y=15$ or $7 x-y=5$.