Part (i):

To find the gradient of line \(AB\), use the formula:

\(m = \frac{y_2 - y_1}{x_2 - x_1}\)

Substitute the coordinates of \(A(3a, -a)\) and \(B(-a, 2a)\):

\(m = \frac{2a - (-a)}{-a - 3a} = \frac{3a}{-4a} = -\frac{3}{4}\)

The equation of the line through the origin parallel to \(AB\) is:

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

Part (ii):

Use the distance formula to find the length of \(AB\):

\(AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Substitute the coordinates:

\(AB = \sqrt{(-a - 3a)^2 + (2a - (-a))^2} = \sqrt{(-4a)^2 + (3a)^2}\)

\(AB = \sqrt{16a^2 + 9a^2} = \sqrt{25a^2} = 5a\)

Given \(AB = 3\frac{1}{3} = \frac{10}{3}\), set up the equation:

\(5a = \frac{10}{3}\)

Solve for \(a\):

\(a = \frac{10}{3} \times \frac{1}{5} = \frac{2}{3}\)