Equation of line passing through the point of intersection of lines $3 x-4 y+1=0$ and $5 x+y-1=0$ is
$$
\begin{aligned}
& (3 x-4 y+1)+\lambda(5 x+y-1)=0 \\
& \Rightarrow(3+5 \lambda) x+(-4+\lambda) y+1-\lambda=0
\end{aligned}
$$
Given line has equal intercepts
$$
\begin{aligned}
& \therefore & 3+5 \lambda & =-4+\lambda \\
\Rightarrow & & \lambda & =-\frac{7}{4}
\end{aligned}
$$
$\therefore$ Equation of line is
$$
23 x+23 y=11 \Rightarrow \frac{x}{\frac{11}{23}}+\frac{y}{\frac{11}{23}}=1
$$
$\therefore$ Area of triangle formed by the line and
$$
\begin{aligned}
\text { coordinates axes } & =\frac{1}{2} \times \frac{11}{23} \times \frac{11}{23} \\
& =\frac{121}{1058}
\end{aligned}
$$