The gradient of OB is calculated as follows:

Gradient of OB = \(\frac{1 - 0}{3 - 0} = \frac{1}{3}\).

Since L₁ is parallel to OB, it has the same gradient: \(\frac{1}{3}\).

The equation of L₁ passing through A (-1, 3) is:

\(y - 3 = \frac{1}{3}(x + 1)\).

The gradient of AB is:

Gradient of AB = \(\frac{1 - 3}{3 + 1} = -\frac{1}{2}\).

The gradient of L₂, being perpendicular to AB, is the negative reciprocal:

\(Gradient of L₂ = 2.\)

The equation of L₂ passing through B (3, 1) is:

\(y - 1 = 2(x - 3)\).

Solving the simultaneous equations:

1. \(y = \frac{1}{3}x + \frac{10}{3}\)

2. \(y = 2x - 5\)

Equating the two expressions for \(y\):

\(\frac{1}{3}x + \frac{10}{3} = 2x - 5\)

Multiply through by 3 to clear the fraction:

\(x + 10 = 6x - 15\)

Rearrange to solve for \(x\):

\(5x = 25\)

\(x = 5\)

Substitute \(x = 5\) back into one of the equations to find \(y\):

\(y = 2(5) - 5 = 5\)

Thus, the coordinates of C are (5, 5).