To determine for what integral value of \(k\) the system of equations has no solution, we first analyze the given set of equations:

1. The first equation is: \(kx - 5y + 6 = 0\) which can be rewritten in the form \(y = \frac{k}{5}x + \frac{6}{5}\).
2. The second equation is: \(4(k-1)y - 12x + 3 = 0\) which can be rewritten in the form \(y = \frac{12}{4(k-1)}x - \frac{3}{4(k-1)}\).

For the system of equations to have no solution, their slopes must be equal and their intercepts must be different, which means the lines are parallel but not coincident.

Equating the slopes:

• From the first equation, the slope is \(\frac{k}{5}\).
• From the second equation, the slope is \(\frac{12}{4(k-1)} = \frac{3}{k-1}\).

Equate the slopes for the lines to be parallel:

\(\frac{k}{5} = \frac{3}{k-1}\)

Cross multiplying gives:

\(k(k-1) = 15\)

Simplifying the quadratic equation:

\(k^2 - k - 15 = 0\)

The quadratic can be solved using the quadratic formula:

\(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Substitute \(a = 1\), \(b = -1\), and \(c = -15\):

\(k = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-15)}}{2(1)} = \frac{1 \pm \sqrt{1 + 60}}{2} = \frac{1 \pm \sqrt{61}}{2}\)

The solutions involve a square root that is not an integer, indicating no intelligible integer solution for \(k\) exists in the form making the lines parallel but distinct.

Therefore, the correct answer is No such value exists since there is no integer solution for which the system of equations will be parallel but not coincident.