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

Let the required line be denoted by $$L$$.
Its symmetric form is given in the question as
$$\frac{x-1}{-2}=\frac{y+4}{d}=\frac{z-c}{-4}\,.$$

Therefore,

 • a point on $$L$$ is $$A(1,\,-4,\,c)$$,
 • its direction vector is $$\vec{n}=(-2,\,d,\,-4)\,.$$

The line $$L$$ is perpendicular to both of the following lines:

$$L_1:\;\vec{r}=\lambda(\hat{i}+a\hat{j}+b\hat{k})\quad\Longrightarrow\quad \text{direction vector }\vec{v_1}=(1,\,a,\,b)$$

$$L_2:\;\vec{r}=(\hat{i}-\hat{j}-6\hat{k})+\mu(-b\hat{i}+a\hat{j}+5\hat{k})\quad\Longrightarrow\quad \text{direction vector }\vec{v_2}=(-b,\,a,\,5)$$

Because $$L$$ is perpendicular to $$L_1$$ and $$L_2$$,

$$\vec{n}\cdot\vec{v_1}=0\quad\text{and}\quad\vec{n}\cdot\vec{v_2}=0\,.$$

Compute each dot product:

1. $$\vec{n}\cdot\vec{v_1}=(-2,\,d,\,-4)\cdot(1,\,a,\,b) =-2+da-4b=0$$ $$\Longrightarrow\;da-4b=2\;-\;(1)$$

2. $$\vec{n}\cdot\vec{v_2}=(-2,\,d,\,-4)\cdot(-b,\,a,\,5) =2b+da-20=0$$ $$\Longrightarrow\;da+2b=20\;-\;(2)$$

The required line $$L$$ must also pass through the point
$$P\Bigl(0,\;-\frac12,\;0\Bigr)$$ given in the statement.

Let the common ratio in the symmetric form be $$t$$. Using point $$A(1,-4,c)$$ and direction $$\vec{n}$$, the parametric equations of $$L$$ are

$$x=1-2t,\qquad y=-4+dt,\qquad z=c-4t\,.$$

Insert point $$P$$:

$$1-2t=0\;\Longrightarrow\;t=\frac12$$

$$-4+dt=-\frac12\;\Longrightarrow\;d\left(\frac12\right)=\frac72 \;\Longrightarrow\;d=7$$

$$c-4t=0\;\Longrightarrow\;c-4\left(\frac12\right)=0 \;\Longrightarrow\;c=2$$

Substitute $$d=7$$ into equations $$(1)$$ and $$(2)$$:

From $$(1):\;7a-4b=2$$

From $$(2):\;7a+2b=20$$

Solve the simultaneous equations:

Add the two equations:
$$\bigl(7a+2b\bigr)+\bigl(7a-4b\bigr)=20+2 \;\Longrightarrow\;14a-2b=22\;-\;(3)$$

Subtract $$(1)$$ from $$(2):\;(7a+2b)-(7a-4b)=20-2 \;\Longrightarrow\;6b=18 \;\Longrightarrow\;b=3$$

Insert $$b=3$$ into $$(1):\;7a-12=2 \;\Longrightarrow\;7a=14 \;\Longrightarrow\;a=2$$

Now we have
$$a=2,\;b=3,\;c=2,\;d=7.$$

Finally,

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

Hence, the value of $$a+b+c+d$$ is $$14$$, which matches Option B.