We know that, distance from point $\left(x_1, y_1\right)$ to the line $a x+b y+c=0$ is
$$
\left|\frac{a x_1+b y_1+c}{\sqrt{a^2+b^2}}\right|
$$
$\therefore$ Distance from $(1,1)$ to $3 x+4 y+c=0$ is 7
$$
\begin{array}{rlr|}
& \quad 7 & =\left|\frac{3(1)+4(1)+c}{\sqrt{3^2+4^2}}\right| \\
\Rightarrow \quad 7 & =\left|\frac{7+c}{5}\right| \\
\Rightarrow \quad 35 & =|7+c| \\
\Rightarrow & 7+c & = \pm 35 \\
\Rightarrow \quad c & =-7 \pm 35 \\
& c & =28,-42
\end{array}
$$