1. Key Concepts and Formulas

   • Position Vector of a Point: If a point P has coordinates $(x, y, z)$, its position vector is $\overrightarrow{OP} = x\widehat i + y\widehat j + z\widehat k$.
   • Vector between Two Points: If A is $(x_1, y_1, z_1)$ and B is $(x_2, y_2, z_2)$, then the vector $\overrightarrow{AB}$ is given by $\overrightarrow{AB} = (x_2 - x_1)\widehat i + (y_2 - y_1)\widehat j + (z_2 - z_1)\widehat k$.
   • Normal Vector of a Plane: For a plane with the equation $ax + by + cz = d$, its normal vector is $\overrightarrow n = a\widehat i + b\widehat j + c\widehat k$.
   • Vector Parallel to a Plane: A vector $\overrightarrow v$ is parallel to a plane if and only if it is perpendicular to the plane's normal vector $\overrightarrow n$. Mathematically, this means their dot product is zero: $\overrightarrow v \cdot \overrightarrow n = 0$.

2. Step-by-Step Solution

   Step 1: Determine the coordinates of point A and point B.

   • What we are doing: We identify the coordinates of point A from the given line equation and note down the coordinates of point B.
   • Why we are doing it: These coordinates are essential for calculating the vector $\overrightarrow{AB}$. Point A lies on the line given by the vector equation: $\overrightarrow r = \left( {1 - 3\mu } \right)\widehat i + \left( {\mu - 1} \right)\widehat j + \left( {2 + 5\mu } \right)\widehat k$ By comparing this with the general form $\overrightarrow r = x\widehat i + y\widehat j + z\widehat k$, the coordinates of point A are: $A = (1 - 3\mu, \mu - 1, 2 + 5\mu)$ Point B is given as: $B = (3, 2, 6)$

   Step 2: Calculate the vector $\overrightarrow{AB}$.

   • What we are doing: We find the component form of the vector connecting point A to point B.
   • Why we are doing it: This is the vector whose parallelism with the plane needs to be checked, and it will be used in the dot product condition. The vector $\overrightarrow{AB}$ is found by subtracting the coordinates of A from the coordinates of B: $\overrightarrow{AB} = (x_B - x_A)\widehat i + (y_B - y_A)\widehat j + (z_B - z_A)\widehat k$ $\overrightarrow{AB} = (3 - (1 - 3\mu))\widehat i + (2 - (\mu - 1))\widehat j + (6 - (2 + 5\mu))\widehat k$ Simplify the components: $\overrightarrow{AB} = (3 - 1 + 3\mu)\widehat i + (2 - \mu + 1)\widehat j + (6 - 2 - 5\mu)\widehat k$ $\overrightarrow{AB} = (2 + 3\mu)\widehat i + (3 - \mu)\widehat j + (4 - 5\mu)\widehat k$

   Step 3: Identify the normal vector to the plane.

   • What we are doing: We extract the coefficients of x, y, and z from the plane's equation to form its normal vector.
   • Why we are doing it: The normal vector is perpendicular to every vector lying in the plane, which is crucial for applying the parallelism condition. The equation of the plane is given as: $x - 4y + 3z = 1$ Comparing this with the general form $ax + by + cz = d$, the normal vector $\overrightarrow n$ is composed of the coefficients of $x$, $y$, and $z$: $\overrightarrow n = 1\widehat i - 4\widehat j + 3\widehat k$

   Step 4: Apply the condition for $\overrightarrow{AB}$ to be parallel to the plane.

   • What we are doing: We set the dot product of $\overrightarrow{AB}$ and $\overrightarrow n$ to zero.
   • Why we are doing it: This is the fundamental mathematical condition for a vector to be parallel to a plane. Since vector $\overrightarrow{AB}$ is parallel to the plane, it must be perpendicular to the plane's normal vector $\overrightarrow n$. Therefore, their dot product must be zero: $\overrightarrow{AB} \cdot \overrightarrow n = 0$ Substitute the expressions for $\overrightarrow{AB}$ and $\overrightarrow n$: $\left( (2 + 3\mu)\widehat i + (3 - \mu)\widehat j + (4 - 5\mu)\widehat k \right) \cdot \left( 1\widehat i - 4\widehat j + 3\widehat k \right) = 0$ Perform the dot product: $(2 + 3\mu)(1) + (3 - \mu)(-4) + (4 - 5\mu)(3) = 0$

   Step 5: Solve the equation for $\mu$.

   • What we are doing: We simplify the equation obtained in Step 4 and isolate $\mu$.
   • Why we are doing it: This step yields the specific value of $\mu$ that satisfies the given condition. Expand and simplify the equation: $2 + 3\mu - 12 + 4\mu + 12 - 15\mu = 0$ Combine the constant terms and the $\mu$ terms: $(2 - 12 + 12) + (3\mu + 4\mu - 15\mu) = 0$ $2 + (7\mu - 15\mu) = 0$ $2 - 8\mu = 0$ Now, solve for $\mu$: $8\mu = 2$ $\mu = \frac{2}{8}$ $\mu = \frac{1}{4}$

3. Common Mistakes & Tips

   • Sign Errors: Be extremely careful with signs when calculating vector components (e.g., $x_B - x_A$) and when performing the dot product, especially with negative coefficients in the normal vector.
   • Incorrect Normal Vector: Ensure the normal vector is correctly extracted from the plane equation. The coefficients of $x, y, z$ directly form the components of the normal vector.
   • Misunderstanding Parallelism: Remember that a vector parallel to a plane is perpendicular to the plane's normal vector. This is why their dot product is zero. Don't confuse it with vectors lying in the plane.

4. Summary

   To find the value of $\mu$ for which vector $\overrightarrow{AB}$ is parallel to the given plane, we first determined the coordinates of point A in terms of $\mu$ and point B. Then, we calculated the vector $\overrightarrow{AB}$. Next, we identified the normal vector of the plane from its equation. Finally, we applied the condition that a vector parallel to a plane must be perpendicular to its normal vector, meaning their dot product is zero. Solving the resulting linear equation for $\mu$ yielded the required value. The calculation resulted in $\mu = \frac{1}{4}$.

5. Final Answer

The final answer is $\boxed{\frac{1}{4}}$ which corresponds to option (C).