Let `P(3,2,6)` be a point in space and `Q` be a point on line ` vec r=( hat i- hat j+2 hat k)+mu(-3 hat i+ hat j+5 hat k)dot` Then the value of `mu` for which the vector ` vec P Q` is parallel to the plane `x-4y+3z=1` is a. 1/4 b. -1/4 c. 1/8 d. -1/8

Options

A. `1//4`
B. `-1//4`
C. `1//8`
D. `-1//8`

Explanation

To solve the problem step by step, we will first determine the coordinates of point Q on the given line and then find the vector PQ. Finally, we will use the condition for parallelism with the given plane to find the value of mu.

### Step 1: Define the points
Let \( P(3, 2, 6) \) be the point in space. The point \( Q \) lies on the line defined by the equation:
\[ \vec{r} = \hat{i} - \hat{j} + 2\hat{k} + \mu(-3\hat{i} + \hat{j} + 5\hat{k}). \]
From this, we can express the coordinates of point \( Q \):
\[ Q = (1 - 3\mu, -1 + \mu, 2 + 5\mu). \]

### Step 2: Find the vector \( \vec{PQ} \)
The vector \( \vec{PQ} \) can be found using the coordinates of points \( P \) and \( Q \):
\[ \vec{PQ} = Q - P = \left( (1 - 3\mu) - 3, (-1 + \mu) - 2, (2 + 5\mu) - 6 \right). \]
This simplifies to:
\[ \vec{PQ} = (-2 - 3\mu, -3 + \mu, -4 + 5\mu). \]

### Step 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 this plane is:
\[ \vec{n} = (1, -4, 3). \]

### Step 4: Use the condition for parallelism
For the vector \( \vec{PQ} \) to be parallel to the plane, it must be perpendicular to the normal vector \( \vec{n} \). This means that their dot product must be zero:
\[ \vec{PQ} \cdot \vec{n} = 0. \]
Calculating the dot product:
\[ (-2 - 3\mu)(1) + (-3 + \mu)(-4) + (-4 + 5\mu)(3) = 0. \]
Expanding this gives:
\[ -2 - 3\mu + 12 - 4\mu - 12 + 15\mu = 0. \]
Combining like terms:
\[ (15\mu - 3\mu - 4\mu) + (-2 + 12 - 12) = 0, \]
which simplifies to:
\[ 8\mu - 2 = 0. \]

### Step 5: Solve for \( \mu \)
Now, we can solve for \( \mu \):
\[ 8\mu = 2 \implies \mu = \frac{2}{8} = \frac{1}{4}. \]

### Conclusion
Thus, the value of \( \mu \) for which the vector \( \vec{PQ} \) is parallel to the plane \( x - 4y + 3z = 1 \) is:
\[ \mu = \frac{1}{4}. \]