Here's a detailed and well-structured solution to the problem, adhering to the specified format and rules.

1. Key Concepts and Formulas

   • Equation of a Line in Cartesian Form: A line passing through a point $P_1(x_1, y_1, z_1)$ with direction ratios $(a_1, b_1, c_1)$ is given by: $\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}$
   • Condition for Coplanarity of Two Lines: Two lines, $L_1$ and $L_2$, are coplanar if they lie in the same plane. This occurs if they are parallel or if they intersect. Let $L_1$ pass through $P_1(x_1, y_1, z_1)$ with direction ratios $(a_1, b_1, c_1)$. Let $L_2$ pass through $P_2(x_2, y_2, z_2)$ with direction ratios $(a_2, b_2, c_2)$. The lines $L_1$ and $L_2$ are coplanar if and only if the scalar triple product of the vector $\vec{P_1P_2}$ and their direction vectors $\vec{d_1}$ and $\vec{d_2}$ is zero. This is expressed using the determinant: $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$
   • Interpretation of the Determinant: The first row represents the vector connecting a point on $L_1$ to a point on $L_2$. The second and third rows represent the direction vectors of $L_1$ and $L_2$, respectively. If these three vectors are coplanar, the lines themselves must be coplanar.

2. Step-by-Step Solution

   Step 1: Identify the point and direction ratios for the given line ($L_1$). The given line is: $L_1: \frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$ Comparing this with the standard form $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$:

   • Point $P_1(x_1, y_1, z_1) = (-3, 1, 5)$.
   • Direction ratios $\vec{d_1}(a_1, b_1, c_1) = (-3, 1, 5)$.

   Step 2: Test each option for coplanarity with $L_1$. For each option, we'll extract a point $P_2(x_2, y_2, z_2)$ and direction ratios $\vec{d_2}(a_2, b_2, c_2)$, then calculate the determinant.

   Option (A): The line is $L_A: \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{4}$

   • Point $P_A(-1, 2, 5)$.
   • Direction ratios $\vec{d_A}(-1, 2, 4)$.
   • Vector $\vec{P_1P_A} = (x_2-x_1, y_2-y_1, z_2-z_1) = (-1 - (-3), 2 - 1, 5 - 5) = (2, 1, 0)$.
   • Calculate the determinant: $\begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 4 \end{vmatrix}$ Expanding along the first row: = 2 \left| \begin{matrix} 1 & 5 \\ 2 & 4 \end{vmatrix} \right| - 1 \left| \begin{matrix} -3 & 5 \\ -1 & 4 \end{vmatrix} \right| + 0 \left| \begin{matrix} -3 & 1 \\ -1 & 2 \end{vmatrix} \right| $= 2((1)(4) - (5)(2)) - 1((-3)(4) - (5)(-1)) + 0$ $= 2(4 - 10) - 1(-12 + 5)$ $= 2(-6) - 1(-7)$ $= -12 + 7 = -5$ Since the determinant is $-5 \neq 0$, line $L_A$ is not coplanar with $L_1$.

   Option (B): The line is $L_B: \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$

   • Point $P_B(-1, 2, 5)$.
   • Direction ratios $\vec{d_B}(-1, 2, 5)$.
   • Vector $\vec{P_1P_B} = (-1 - (-3), 2 - 1, 5 - 5) = (2, 1, 0)$.
   • Calculate the determinant: $\begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}$ Expanding along the first row: = 2 \left| \begin{matrix} 1 & 5 \\ 2 & 5 \end{vmatrix} \right| - 1 \left| \begin{matrix} -3 & 5 \\ -1 & 5 \end{vmatrix} \right| + 0 \left| \begin{matrix} -3 & 1 \\ -1 & 2 \end{vmatrix} \right| $= 2((1)(5) - (5)(2)) - 1((-3)(5) - (5)(-1)) + 0$ $= 2(5 - 10) - 1(-15 + 5)$ $= 2(-5) - 1(-10)$ $= -10 + 10 = 0$ Since the determinant is $0$, line $L_B$ is coplanar with $L_1$.

   Option (C): The line is $L_C: \frac{x-1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$

   • Point $P_C(1, 2, 5)$.
   • Direction ratios $\vec{d_C}(-1, 2, 5)$.
   • Vector $\vec{P_1P_C} = (1 - (-3), 2 - 1, 5 - 5) = (4, 1, 0)$.
   • Calculate the determinant: $\begin{vmatrix} 4 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}$ Expanding along the first row: $= 4((1)(5) - (5)(2)) - 1((-3)(5) - (5)(-1)) + 0$ $= 4(5 - 10) - 1(-15 + 5)$ $= 4(-5) - 1(-10)$ $= -20 + 10 = -10$ Since the determinant is $-10 \neq 0$, line $L_C$ is not coplanar with $L_1$.

   Option (D): The line is $L_D: \frac{x+1}{1}=\frac{y-2}{2}=\frac{z-5}{5}$

   • Point $P_D(-1, 2, 5)$.
   • Direction ratios $\vec{d_D}(1, 2, 5)$.
   • Vector $\vec{P_1P_D} = (-1 - (-3), 2 - 1, 5 - 5) = (2, 1, 0)$.
   • Calculate the determinant: $\begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ 1 & 2 & 5 \end{vmatrix}$ Expanding along the first row: $= 2((1)(5) - (5)(2)) - 1((-3)(5) - (5)(1)) + 0$ $= 2(5 - 10) - 1(-15 - 5)$ $= 2(-5) - 1(-20)$ $= -10 + 20 = 10$ Since the determinant is $10 \neq 0$, line $L_D$ is not coplanar with $L_1$.

3. Common Mistakes & Tips

   • Sign Errors in Point Extraction: Always remember that the standard form is $\frac{x-x_1}{a_1}$. So, if you have $(x+3)$, $x_1$ is $-3$, not $3$.
   • Determinant Calculation: Be meticulous with signs during determinant expansion, especially when subtracting terms. A common mistake is $-(a \cdot d - b \cdot c)$ becoming $-ad - bc$ instead of $-ad + bc$.
   • Parallel Lines: If two lines are parallel, their direction ratios will be proportional, meaning $(a_1, b_1, c_1) = k(a_2, b_2, c_2)$ for some scalar $k$. In this case, the second and third rows of the coplanarity determinant will be proportional, making the determinant automatically zero. Thus, parallel lines are always coplanar. Check for this shortcut first if the direction vectors seem related. (In this problem, no options were parallel to $L_1$).

4. Summary

   To determine which line is coplanar with the given line, we applied the condition for coplanarity, which involves setting a specific determinant to zero. We extracted a point and direction vector from the given line and then did the same for each option. For each option, we formed a $3 \times 3$ determinant using the vector connecting the two points and their respective direction vectors. A determinant value of zero indicates coplanarity. Our calculations showed that only Option (B) satisfied this condition, yielding a determinant of 0.

The final answer is $\boxed{A}$.