Find the coordinates of N, given A(6; -2), B(2; 15), C(-4; 3), and M is the midpoint of AB and MN || BC.

The midpoint formula for two points $(x_1, y_1)$ and $(x_2, y_2)$ is $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
Given $A(6; -2)$ and $B(2; 15)$.

$$M = \left(\frac{6+2}{2}, \frac{-2+15}{2}\right)$$ $$M = \left(\frac{8}{2}, \frac{13}{2}\right)$$ $$M = \left(4, \frac{13}{2}\right)$$

The coordinates of M are $\boxed{\left(4, \frac{13}{2}\right)}$.

Since $MN \parallel BC$, their gradients are equal. The gradient formula for two points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2-y_1}{x_2-x_1}$.
Given $B(2; 15)$ and $C(-4; 3)$.

$$m_{BC} = \frac{3-15}{-4-2}$$ $$m_{BC} = \frac{-12}{-6}$$ $$m_{BC} = 2$$

Therefore, the gradient of line MN is $\boxed{2}$.

We have the gradient $m_{MN} = 2$ (from Step 2) and point $M\left(4, \frac{13}{2}\right)$ (from Step 1).
Using the point-slope form $y - y_1 = m(x - x_1)$:

$$y - \frac{13}{2} = 2(x - 4)$$ $$y - \frac{13}{2} = 2x - 8$$ $$y = 2x - 8 + \frac{13}{2}$$ $$y = 2x - \frac{16}{2} + \frac{13}{2}$$ $$y = 2x - \frac{3}{2}$$

The equation of line MN is $\boxed{y = 2x - \frac{3}{2}}$.

Given that $M$ is the midpoint of $AB$ and $MN \parallel BC$.
By the Midpoint Theorem (or its converse), if a line passes through the midpoint of one side of a triangle and is parallel to another side, then it bisects the third side.
Therefore, $N$ is the midpoint of $AC$.
Given $A(6; -2)$ and $C(-4; 3)$.

$$N = \left(\frac{6+(-4)}{2}, \frac{-2+3}{2}\right)$$ $$N = \left(\frac{2}{2}, \frac{1}{2}\right)$$ $$N = \left(1, \frac{1}{2}\right)$$

The coordinates of N are $\boxed{\left(1, \frac{1}{2}\right)}$.

In a parallelogram, the diagonals bisect each other. This means the midpoint of $AC$ is the same as the midpoint of $BD$.
From Step 4, the midpoint of $AC$ is $N\left(1, \frac{1}{2}\right)$.
Let $D = (x_D, y_D)$. The midpoint of $BD$ is $\left(\frac{x_B+x_D}{2}, \frac{y_B+y_D}{2}\right)$.
Given $B(2; 15)$.
Equating the midpoints:

$$\frac{2+x_D}{2} = 1 \implies 2+x_D = 2 \implies x_D = 0$$ $$\frac{15+y_D}{2} = \frac{1}{2} \implies 15+y_D = 1 \implies y_D = 1-15 \implies y_D = -14$$

The coordinates of D are $\boxed{(0, -14)}$.