1. Key Concepts and Formulas

• Equation of a Line in 3D: A line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(l, m, n)$ can be represented in Cartesian form as $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$. The vector $\vec{v} = l\hat{i} + m\hat{j} + n\hat{k}$ is the direction vector of the line.
• Direction Vector of Perpendicular Lines: If a line $L$ is perpendicular to two other lines $L_1$ and $L_2$ with direction vectors $\vec{v_1}$ and $\vec{v_2}$ respectively, then the direction vector of $L$, let's call it $\vec{v}$, must be parallel to the cross product of $\vec{v_1}$ and $\vec{v_2}$. That is, $\vec{v} = k (\vec{v_1} \times \vec{v_2})$ for some non-zero scalar $k$.
• Point on a Line: If a point $(x_0, y_0, z_0)$ lies on a line, its coordinates must satisfy the equation of the line.

2. Step-by-Step Solution

Step 1: Identify the direction vectors of the lines to which the required line is perpendicular. The required line is perpendicular to two given lines:

1. $L_1: \vec{r} = \lambda \left( \hat{i} + a\hat{j} + b\hat{k} \right)$. The direction vector of $L_1$ is $\vec{v_1} = \hat{i} + a\hat{j} + b\hat{k}$.
2. $L_2: \vec{r} = \left( \hat{i} - \hat{j} - 6\hat{k} \right) + \mu \left( -b \hat{i} + a\hat{j} + 5\hat{k} \right)$. The direction vector of $L_2$ is $\vec{v_2} = -b \hat{i} + a\hat{j} + 5\hat{k}$.

Step 2: Determine the direction vector of the required line. Since the required line $L$ is perpendicular to both $L_1$ and $L_2$, its direction vector, $\vec{v}$, must be parallel to the cross product of $\vec{v_1}$ and $\vec{v_2}$. We calculate this cross product: $\vec{v} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & a & b \\ -b & a & 5 \end{vmatrix}$ Expanding the determinant: $\vec{v} = \hat{i}(a \cdot 5 - b \cdot a) - \hat{j}(1 \cdot 5 - b \cdot (-b)) + \hat{k}(1 \cdot a - a \cdot (-b))$ $\vec{v} = (5a - ab)\hat{i} - (5 + b^2)\hat{j} + (a + ab)\hat{k}$ The direction ratios of the required line $L$ are therefore proportional to $(5a - ab, -(b^2 + 5), a + ab)$.

Step 3: Relate the calculated direction ratios to the given equation of the required line. The equation of the required line $L$ is given as $\frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4}$. From this equation, the direction ratios of the line are $(-2, d, -4)$.

Since the direction ratios from the cross product must be proportional to these given direction ratios, we can set up the following proportionality: $\frac{5a - ab}{-2} = \frac{-(b^2 + 5)}{d} = \frac{a + ab}{-4} = k$ where $k$ is a constant of proportionality. This implies:

1. $5a - ab = -2k$
2. $-(b^2 + 5) = dk$
3. $a + ab = -4k$

Step 4: Use the given point on the line to find the values of $c$ and $d$. The required line $L$ passes through the point $P \left( 0, -\frac{1}{2}, 0 \right)$. We substitute these coordinates into the line's equation: $\frac{0-1}{-2} = \frac{-\frac{1}{2}+4}{d} = \frac{0-c}{-4}$ Simplify each part:

• $\frac{-1}{-2} = \frac{1}{2}$
• $\frac{-\frac{1}{2}+4}{d} = \frac{\frac{-1+8}{2}}{d} = \frac{\frac{7}{2}}{d} = \frac{7}{2d}$
• $\frac{-c}{-4} = \frac{c}{4}$

Now, equate these simplified expressions: $\frac{1}{2} = \frac{7}{2d} = \frac{c}{4}$ From $\frac{1}{2} = \frac{7}{2d}$: $2d = 14 \implies d = 7$

From $\frac{1}{2} = \frac{c}{4}$: $2c = 4 \implies c = 2$ So, we have found $c=2$ and $d=7$.

Step 5: Solve for the values of $a$ and $b$. Substitute $d=7$ into the proportionality equations from Step 3: $\frac{5a - ab}{-2} = \frac{-(b^2 + 5)}{7} = \frac{a + ab}{-4}$ We can use the first and third terms to solve for $a$ and $b$: $\frac{5a - ab}{-2} = \frac{a + ab}{-4}$ Multiply both sides by $-4$: $2(5a - ab) = a + ab$ $10a - 2ab = a + ab$ Rearrange terms: $9a - 3ab = 0$ Factor out $3a$: $3a(3 - b) = 0$ This equation implies either $3a = 0$ or $3 - b = 0$. If $a=0$, the direction ratios from Step 2 would be $(0, -(b^2+5), 0)$. For this to be proportional to $(-2, d, -4)$, it would imply $-(b^2+5)=0$, or $b^2=-5$, which has no real solution for $b$. Therefore, $a \neq 0$. Thus, we must have $3 - b = 0$, which gives $b = 3$.

Now substitute $b=3$ into the proportionality, using the second and third terms: $\frac{-(b^2 + 5)}{7} = \frac{a + ab}{-4}$ Substitute $b=3$: $\frac{-((3)^2 + 5)}{7} = \frac{a + a(3)}{-4}$ $\frac{-(9 + 5)}{7} = \frac{4a}{-4}$ $\frac{-14}{7} = -a$ $-2 = -a$ This gives $a = 2$. So, we have found $a=2$ and $b=3$.

Step 6: Calculate the final sum $a+b+c+d$. We have determined the values of all the unknowns:

• $a = 2$
• $b = 3$
• $c = 2$
• $d = 7$

Now, sum these values: $a+b+c+d = 2 + 3 + 2 + 7 = 14$

3. Common Mistakes & Tips

• Algebraic Errors in Cross Product: Be meticulous when calculating the cross product determinant. Incorrect signs or terms are common pitfalls.
• Proportionality: Remember that direction ratios are proportional, not necessarily equal. Using a constant of proportionality (like $k$) or correctly cross-multiplying proportional terms is crucial.
• Checking Special Cases: When solving equations like $3a(3-b)=0$, always consider all possible solutions ($a=0$ or $b=3$) and check if any lead to contradictions with other information in the problem (e.g., $a=0$ led to $b^2=-5$, which is not possible).
• Systematic Substitution: As soon as a variable's value is found (like $c$ and $d$), substitute it back into subsequent equations. This simplifies the remaining calculations and reduces the number of unknowns.

4. Summary This problem required a multi-step approach involving key concepts from 3D geometry. We first determined the direction vector of the required line by taking the cross product of the direction vectors of the two lines it was perpendicular to. Then, by substituting the coordinates of the given point into the line's equation, we found the values of $c$ and $d$. Finally, we equated the components of the direction vectors (using the constant of proportionality) and solved the resulting system of equations to find $a$ and $b$. All the determined values were then summed to obtain the final answer.

The final answer is $\boxed{14}$, which corresponds to option (B).