`L_(1)andL_(2)` are two lines whose vector equations are
`L_(1):vecr=lamda((costheta+sqrt3)hati+(sqrt2sintheta)hatj+(costheta-sqrt3)hatk)`
`L_(2):vecr=mu(ahati+bhatj+chatk)`, where `lamdaandmu` are scalars and `alpha` is the acute angle between `L_(1)andL_(2)`.If the `anglealpha` is independent of `theta` then the value of `alpha` is

Options

A. `(pi)/(6)`
B. `(pi)/(4)`
C. `(pi)/(3)`
D. `(pi)/(2)`

Explanation

To solve the problem, we need to find the acute angle \( \alpha \) between the two lines \( L_1 \) and \( L_2 \) given their vector equations. The angle \( \alpha \) is independent of \( \theta \).

### Step-by-Step Solution:

1. **Identify the Direction Ratios**:
The vector equations of the lines are given as:
\[ L_1: \vec{r} = \lambda \left( \cos \theta + \sqrt{3} \hat{i} + \sqrt{2} \sin \theta \hat{j} + \cos \theta - \sqrt{3} \hat{k} \right) \]
The direction ratios of line \( L_1 \) can be extracted as:
\[ \vec{d_1} = \left( \cos \theta + \sqrt{3}, \sqrt{2} \sin \theta, \cos \theta - \sqrt{3} \right) \]

For line \( L_2 \):
\[ L_2: \vec{r} = \mu (a \hat{i} + b \hat{j} + c \hat{k}) \]
The direction ratios of line \( L_2 \) are:
\[ \vec{d_2} = (a, b, c) \]

2. **Use the Cosine Formula for Angle Between Two Vectors**:
The cosine of the angle \( \alpha \) between the two lines can be expressed as:
\[ \cos \alpha = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|} \]

3. **Calculate the Dot Product**:
The dot product \( \vec{d_1} \cdot \vec{d_2} \) is:
\[ \vec{d_1} \cdot \vec{d_2} = (a(\cos \theta + \sqrt{3}) + b(\sqrt{2} \sin \theta) + c(\cos \theta - \sqrt{3})) \]
Simplifying this gives:
\[ \vec{d_1} \cdot \vec{d_2} = (a + c) \cos \theta + b \sqrt{2} \sin \theta + (a - c) \sqrt{3} \]

4. **Calculate the Magnitudes**:
The magnitude of \( \vec{d_1} \) is:
\[ |\vec{d_1}| = \sqrt{(\cos \theta + \sqrt{3})^2 + (\sqrt{2} \sin \theta)^2 + (\cos \theta - \sqrt{3})^2} \]
Expanding this:
\[ |\vec{d_1}| = \sqrt{(1 + 3 + 2\sqrt{3} \cos \theta) + 2 \sin^2 \theta + (1 + 3 - 2\sqrt{3} \cos \theta)} = \sqrt{4 + 2 + 2\sqrt{3} \cos \theta} \]
Simplifying gives:
\[ |\vec{d_1}| = \sqrt{8 + 2\sqrt{3} \cos \theta} \]

The magnitude of \( \vec{d_2} \) is:
\[ |\vec{d_2}| = \sqrt{a^2 + b^2 + c^2} \]

5. **Set Conditions for Independence of \( \theta \)**:
For \( \alpha \) to be independent of \( \theta \), the coefficients of \( \cos \theta \) and \( \sin \theta \) in the expression for \( \cos \alpha \) must equal zero:
- From \( a + c = 0 \) (let's call this Equation 1)
- From \( b = 0 \) (let's call this Equation 2)

6. **Substituting Back**:
Substitute \( c = -a \) and \( b = 0 \) into the expression for \( \cos \alpha \):
\[ \cos \alpha = \frac{\sqrt{3}(a - (-a))}{\sqrt{8} \sqrt{a^2 + 0 + a^2}} = \frac{2\sqrt{3}a}{\sqrt{8} \cdot \sqrt{2a^2}} = \frac{\sqrt{3}}{2} \]

7. **Finding the Angle**:
Since \( \cos \alpha = \frac{\sqrt{3}}{2} \), we find:
\[ \alpha = \frac{\pi}{6} \]

### Final Answer:
Thus, the value of \( \alpha \) is:
\[ \alpha = \frac{\pi}{6} \]