Find the equation of the line perpendicular to the line $3x-2y=6$ and passes through the point (1, 2).

Step 1 ：Step 1: We first find the slope of the given line. The equation of the line is in the standard form $Ax+By=C$. We can convert it to the slope-intercept form $y=mx+b$ by solving for y. $$y=\frac{3x}{2}-3$$ The slope $m$ of the given line is $\frac{3}{2}$.

Step 2 ：Step 2: The slope of the line perpendicular to the given line is the negative reciprocal of the slope of the given line. So, the slope $m^{\prime }$ of the perpendicular line is $$-\frac{1}{m}=-\frac{2}{3}$$.

Step 3 ：Step 3: Now we know the slope of the perpendicular line and a point it passes through (1, 2), we can find the equation of the perpendicular line using the point-slope form $y-y1=m^{\prime }(x-x1)$. Substituting $m^{\prime }=-\frac{2}{3}$, $x1=1$, and $y1=2$ into the equation, we get $$y-2=-\frac{2}{3}(x-1)$$.

Step 4 ：Step 4: Finally, we simplify the above equation to find the equation of the perpendicular line. $$y=-\frac{2}{3}x+\frac{4}{3}+2$$ $$y=-\frac{2}{3}x+\frac{10}{3}$$