First, calculate the gradient of line AB:

Gradient of AB = \(\frac{2 - 8}{6 - 3} = -2\).

Since DC is parallel to AB, the gradient of DC is also -2. Using point C (10, 2), the equation of line DC is:

\(y - 2 = -2(x - 10)\)

Simplifying gives:

\(y = -2x + 20 + 2\)

\(y = -2x + 22\)

For line DA, since it is perpendicular to AB, the gradient of DA is the negative reciprocal of -2, which is \(\frac{1}{2}\). Using point A (3, 8), the equation of line DA is:

\(y - 8 = \frac{1}{2}(x - 3)\)

Simplifying gives:

\(y = \frac{1}{2}x + \frac{1}{2} \times (-3) + 8\)

\(y = \frac{1}{2}x - \frac{3}{2} + 8\)

\(y = \frac{1}{2}x + \frac{13}{2}\)

Now, solve the simultaneous equations:

1. \(y = -2x + 22\)

2. \(y = \frac{1}{2}x + \frac{13}{2}\)

Equating the two expressions for \(y\):

\(-2x + 22 = \frac{1}{2}x + \frac{13}{2}\)

Multiply through by 2 to eliminate the fraction:

\(-4x + 44 = x + 13\)

\(44 - 13 = 4x + x\)

\(31 = 5x\)

\(x = \frac{31}{5} = 6.2\)

Substitute \(x = 6.2\) back into equation 1:

\(y = -2(6.2) + 22\)

\(y = -12.4 + 22\)

\(y = 9.6\)

Thus, the coordinates of D are (6.2, 9.6).