Two points, A (5,2) and B (-3,8), lie on a straight line l and C (-2,4), lies on another straight line m. Find (i) the gradient of l, (ii) AB as a column vector, (iii) the length of the line AB. Given that m is parallel to l, find the equation of m in its simplest form.

The formula for the gradient $m$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

$$m_l = \frac{8 - 2}{-3 - 5} = \frac{6}{-8} = -\frac{3}{4}$$

The gradient of $l$ is $\boxed{-\frac{3}{4}}$.

The column vector from $A(x_A, y_A)$ to $B(x_B, y_B)$ is $\begin{pmatrix} x_B - x_A \\ y_B - y_A \end{pmatrix}$.

$$\vec{AB} = \begin{pmatrix} -3 - 5 \\ 8 - 2 \end{pmatrix} = \begin{pmatrix} -8 \\ 6 \end{pmatrix}$$

$\vec{AB}$ as a column vector is $\boxed{\begin{pmatrix} -8 \\ 6 \end{pmatrix}}$.

The length of a line segment is given by the distance formula $D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Using the components of the vector $\vec{AB}$:

$$\text{Length } AB = \sqrt{(-8)^2 + (6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$$

The length of the line $AB$ is $\boxed{10 \text{ units}}$.

Line $m$ is parallel to line $l$, so its gradient $m_m$ is equal to the gradient of $l$.

$$m_m = m_l = -\frac{3}{4}$$

Line $m$ passes through point $C(-2, 4)$. Using the point-slope form $y - y_1 = m(x - x_1)$:

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

$$y - 4 = -\frac{3}{4}x - \frac{3}{4} \times 2$$ $$y - 4 = -\frac{3}{4}x - \frac{6}{4}$$ $$y - 4 = -\frac{3}{4}x - \frac{3}{2}$$ $$y = -\frac{3}{4}x - \frac{3}{2} + 4$$ $$y = -\frac{3}{4}x - \frac{3}{2} + \frac{8}{2}$$ $$y = -\frac{3}{4}x + \frac{5}{2}$$

The equation of line $m$ in its simplest form is $\boxed{y = -\frac{3}{4}x + \frac{5}{2}}$.