Let A be a point on the line `vec(r)=(1-3mu)hati+(mu-1)hatj+(2+5mu)hatk` and `B(3,2,6)` be a point in the space. Then the value of `mu` for which the vector `vec(AB)` is parallel to the plane `x-4y+3z=1` is: (a) `1/8` (b) `1/2` (c) `1/4` (d) `-1/4`

Explanation

To solve the problem, we need to find the value of \( \mu \) such that the vector \( \vec{AB} \) is parallel to the plane defined by the equation \( x - 4y + 3z = 1 \).

### Step-by-Step Solution:

1. **Identify the Points**:
- The point \( A \) lies on the line given by the vector equation:
\[ \vec{r} = (1 - 3\mu) \hat{i} + (\mu - 1) \hat{j} + (2 + 5\mu) \hat{k} \]
- The coordinates of point \( A \) can be expressed as:
\[ A(1 - 3\mu, \mu - 1, 2 + 5\mu) \]
- The coordinates of point \( B \) are given as:
\[ B(3, 2, 6) \]

2. **Find the Vector \( \vec{AB} \)**:
- The vector \( \vec{AB} \) is calculated as:
\[ \vec{AB} = \vec{B} - \vec{A} = (3 - (1 - 3\mu)) \hat{i} + (2 - (\mu - 1)) \hat{j} + (6 - (2 + 5\mu)) \hat{k} \]
- Simplifying each component:
- \( x \)-component: \( 3 - (1 - 3\mu) = 3 - 1 + 3\mu = 2 + 3\mu \)
- \( y \)-component: \( 2 - (\mu - 1) = 2 + 1 - \mu = 3 - \mu \)
- \( z \)-component: \( 6 - (2 + 5\mu) = 6 - 2 - 5\mu = 4 - 5\mu \)
- Thus, we have:
\[ \vec{AB} = (2 + 3\mu) \hat{i} + (3 - \mu) \hat{j} + (4 - 5\mu) \hat{k} \]

3. **Find the Normal Vector of the Plane**:
- The equation of the plane is given by \( x - 4y + 3z = 1 \).
- The normal vector \( \vec{n} \) to the plane can be identified from the coefficients of \( x, y, z \):
\[ \vec{n} = \hat{i} - 4\hat{j} + 3\hat{k} \]

4. **Condition for Parallelism**:
- For \( \vec{AB} \) to be parallel to the plane, it must be perpendicular to the normal vector \( \vec{n} \). This means:
\[ \vec{AB} \cdot \vec{n} = 0 \]
- Calculating the dot product:
\[ (2 + 3\mu) \cdot 1 + (3 - \mu)(-4) + (4 - 5\mu) \cdot 3 = 0 \]
- Expanding this:
\[ 2 + 3\mu - 12 + 4\mu + 12 - 15\mu = 0 \]
- Combining like terms:
\[ (3\mu + 4\mu - 15\mu) + (2 - 12 + 12) = 0 \]
\[ -8\mu + 2 = 0 \]

5. **Solve for \( \mu \)**:
- Rearranging gives:
\[ -8\mu = -2 \implies \mu = \frac{2}{8} = \frac{1}{4} \]

### Final Answer:
The value of \( \mu \) for which the vector \( \vec{AB} \) is parallel to the plane is:
\[ \mu = \frac{1}{4} \]