First of all, let's find the slope of line L. To find the slope of a line, you write it in the explitic form [tex] y = mx+q [/tex], and consider the coefficient [tex] m [/tex]. So, we have
[tex] 2x - 3y = 5 \iff 3y = 2x-5 \iff y = \dfrac{2}{3} x - \dfrac{5}{3} [/tex]
So, the slope of line L is 2/3.
Two lines are parallel if they have the same slope. So, line M has slope 2/3 as well, and we know that it passes through the point (3, -10).
When you are given the slope [tex] m [/tex] and a point [tex](x_0,y_0)[/tex] of a line, its equation is given by
[tex] y-y_0=m(x-x_0) [/tex]
So, if you plug your values, you have
[tex] y+10 = \dfrac{2}{3}(x-3) [/tex]
which you can simplify into
[tex] y+10 = \dfrac{2}{3}x-2 [/tex]
[tex] y = \dfrac{2}{3}x-12 [/tex]
