Line $\ell_1$ represents the graph of $3x + 2y = -14 - 8x + 5y$. Line $\ell_2$ passes through the point $(5,-6)$, and is perpendicular to line $\ell_1$. Write the equation of line $\ell_2$ in the form $y=mx +b$.
To find the slope of line $\ell_1$, we convert the equation $3x + 2y = -14 - 8x + 5y$ into slope-intercept form, $y = mx + b$. Solving this equation for $y$, we find $y = -11x/3 - 14/3$. Therefore, line $\ell_1$ has slope $-11/3$.
Since line $\ell_2$ is perpendicular to line $\ell_1$, the slope of line $\ell_2$ is the negative reciprocal of the slope of line $\ell_1$. Therefore, the slope of line $\ell_2$ is $3/11$.
To find the equation of line $\ell_2$, we substitute the slope $3/11$ and the point $(5,-6)$ into the point-slope equation $y - y_1 = m(x - x_1)$. We have $y - (-6) = (3/11)(x - 5)$, which simplifies to $y + 6 = (3/11)x - 15/11$. Subtracting $6$ from both sides, we have $y = (3/11)x - 81/11$, so the equation of line $\ell_2$ is $\boxed{y=\frac{3}{11} x - \frac{81}{11}}$.