First, find the midpoint C of line AB:

\(C = \left( \frac{14 + (-6)}{2}, \frac{-7 + 3}{2} \right) = (4, -2)\)

Next, find the slope of AB:

\(m_{AB} = \frac{3 - (-7)}{-6 - 14} = \frac{10}{-20} = -\frac{1}{2}\)

The slope of line CD, being perpendicular to AB, is the negative reciprocal:

\(m_{CD} = 2\)

Using point C(4, -2) in the point-slope form:

\(y + 2 = 2(x - 4)\)

Simplifying to slope-intercept form:

\(y = 2x - 8 - 2\)

\(y = 2x - 10\)

Since the line crosses the y-axis at D, set x = 0:

\(y = 2(0) - 10 = -10\)

So, the equation of CD is:

\(y = 2x - 10\)

For the distance AD, use the distance formula:

\(AD = \sqrt{(14 - 4)^2 + (-7 - (-2))^2}\)

\(AD = \sqrt{10^2 + (-5)^2}\)

\(AD = \sqrt{100 + 25}\)

\(AD = \sqrt{125} = 5\sqrt{5}\)

Therefore, the distance AD is:

\(AD = 5\sqrt{5}\)