The line we need to find should be orthogonal to two given lines, meaning it will be parallel to the cross product of their direction vectors:

$ (\hat{i} + a \hat{j} + b \hat{k}) \times (-b \hat{i} + a \hat{j} + 5 \hat{k}) $

Calculating the cross product results in the vector:

$ \hat{i}(5a - ab) - \hat{j}(b^2 + 5) + \hat{k}(a + ab) $

This shows that the direction ratios of the required line are proportional to:

$ \alpha(5a - ab), - (b^2 + 5), (a + ab) $

Since the line is also expressed as $ \frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4} $, the direction ratios based on this equation are:

$ \alpha(-2), d, -4 $

Thus, we have the equation:

$ \frac{5a - ab}{-2} = \frac{- (b^2 + 5)}{d} = \frac{a + ab}{-4} $

The point $\left(0, -\frac{1}{2}, 0 \right)$ lies on the line, which implies:

$ \frac{0-1}{-2} = \frac{-\frac{1}{2} + 4}{d} = \frac{0-c}{-4} $

Simplifying these gives $ d = 7 $ and $ c = 2 $.

Substitute into the main equation:

$ \frac{5a - ab}{-2} = \frac{a + ab}{-4} $

$ \frac{-b^2 - 5}{7} = \frac{a + ab}{-4} $

Solving these yields:

$ \begin{array}{c|c} -20a + 4ab = -2a - 2ab & 4b^2 + 20 = 70 + 7ab \\ 18a = 6ab & 36 + 20 = 70 + 21a \\ b = 3 & 56 = 28a \Rightarrow a = 2 \end{array} $

Substituting these into the variables, we find:

$ a + b + c + d = 2 + 3 + 2 + 7 = 14 $