If the equation of the line passing through the point $(0, -\tfrac{1}{2}, 0)$ and perpendicular to the lines $\mathbf{r} = \lambda(\hat{i} + a\hat{j} + b\hat{k})$ and $\mathbf{r} = (\hat{i} - \hat{j} - 6\hat{k}) + \mu(-b\hat{i} + a\hat{j} + 5\hat{k})$ is $\frac{x - 1}{-2} = \frac{y + 4}{d} = \frac{z - c}{-4}$, then $a + b + c + d$ is equal to:

Step 1: Find the direction vector of the required line

The required line is perpendicular to both given lines. Therefore, its direction vector must be perpendicular to the direction vectors of both given lines. We can find this common perpendicular direction vector by taking the cross product of the direction vectors of the two given lines. The direction vector of the first line is $\mathbf{v_1} = \hat{i} + a\hat{j} + b\hat{k}$ and the direction vector of the second line is $\mathbf{v_2} = -b\hat{i} + a\hat{j} + 5\hat{k}$.

Step 2: Calculate the cross product

We compute the cross product of the two direction vectors: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & a & b \\ -b & a & 5 \end{vmatrix} = \hat{i}(5a - ab) - \hat{j}(5 - (-b^2)) + \hat{k}(a - (-ab))$$ $$= (5a - ab)\hat{i} - (5 + b^2)\hat{j} + (a + ab)\hat{k}$$

Step 3: Compare direction vectors to find a and b

The direction vector of the required line is proportional to the direction vector given in the equation $\frac{x - 1}{-2} = \frac{y + 4}{d} = \frac{z - c}{-4}$, which is $(-2, d, -4)$. We can equate the ratios of corresponding components. From the first and third components, we have $\frac{5a - ab}{-2} = \frac{a + ab}{-4}$.

Step 4: Solve for a and b

From the equality $\frac{5a - ab}{-2} = \frac{a + ab}{-4}$, we cross-multiply to get $-4(5a - ab) = -2(a + ab)$. Dividing by $-2$ gives $2(5a - ab) = a + ab$. Expanding this, we get $10a - 2ab = a + ab$. Rearranging terms, we have $9a = 3ab$. This implies $3a(3 - b) = 0$. Since $a$ cannot be zero (otherwise the direction vector would be trivial), we must have $3 - b = 0$, which means $b = 3$. Now substitute $b=3$ into the direction vector components: $5a - ab = 5a - 3a = 2a$ and $a + ab = a + 3a = 4a$.

Step 5: Determine d and c

With $b=3$, the direction vector components are $2a$, $-(5+3^2) = -14$, and $4a$. So the direction vector is $(2a, -14, 4a)$. Comparing this with $(-2, d, -4)$, we have $\frac{2a}{-2} = \frac{-14}{d} = \frac{4a}{-4}$. From $\frac{2a}{-2} = \frac{4a}{-4}$, we get $-a = -a$, which is consistent. From $\frac{2a}{-2} = \frac{-14}{d}$, we have $-a = \frac{-14}{d}$, so $ad = 14$. Also, from $\frac{4a}{-4} = \frac{-14}{d}$, we get $-a = \frac{-14}{d}$, which is the same. Since the direction vector components are $(-2, d, -4)$, we can set $2a = -2$, which implies $a = -1$. Substituting $a = -1$ into $ad = 14$, we get $(-1)d = 14$, so $d = -14$. The line passes through $(0, -\frac{1}{2}, 0)$ and its equation is $\frac{x - 1}{-2} = \frac{y + 4}{d} = \frac{z - c}{-4}$. Since $(0, -\frac{1}{2}, 0)$ lies on the line, we substitute these coordinates into the equation: $\frac{0 - 1}{-2} = \frac{-\frac{1}{2} + 4}{d} = \frac{0 - c}{-4}$. This gives $\frac{1}{2} = \frac{\frac{7}{2}}{d} = \frac{-c}{-4}$. From $\frac{1}{2} = \frac{\frac{7}{2}}{d}$, we get $d = 7$. This contradicts $d=-14$. Let's re-evaluate. The direction vector of the required line is proportional to $(-2, d, -4)$. So, $(2a, -14, 4a) = k(-2, d, -4)$ for some scalar $k$. Comparing the first components: $2a = -2k \implies a = -k$. Comparing the third components: $4a = -4k \implies a = -k$. This is consistent. Comparing the second components: $-14 = kd$. Since $a = -k$, we have $k = -a$. So, $-14 = (-a)d \implies ad = 14$. We also know that the point $(0, -\frac{1}{2}, 0)$ lies on the line $\frac{x - 1}{-2} = \frac{y + 4}{d} = \frac{z - c}{-4}$. Substituting the point: $\frac{0 - 1}{-2} = \frac{-\frac{1}{2} + 4}{d} = \frac{0 - c}{-4}$. This simplifies to $\frac{1}{2} = \frac{\frac{7}{2}}{d} = \frac{-c}{-4}$. From $\frac{1}{2} = \frac{\frac{7}{2}}{d}$, we get $d = 7$. Now we have $ad = 14$ and $d = 7$, so $a(7) = 14 \implies a = 2$. From $\frac{1}{2} = \frac{-c}{-4}$, we get $\frac{1}{2} = \frac{c}{4}$, so $c = 2$.

Step 6: Calculate a + b + c + d

We have found the values: $a = 2$, $b = 3$, $c = 2$, and $d = 7$. Now we sum these values.

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