Consider three vectors
`vec(V_(1))=(sin theta)hati+(cos theta)hatj+(a-3)hatk, vec(V_(2))=(sin theta+cos theta)+hati+(cos theta-sin theta)hatj` and `+(b-4)hatk`
`vec(V_(3))=(cos theta)hati+(sin theta)hatj+(c-5)hatk`. If the resultant of `vec(V_(1)),vec(V_(2)) and vec(V_(3))` is equal to `lambda hati`, where `theta in [-pi, pi]` and `a, b, c in N,` then the number of quadruplets `(a, b, c, theta)` are

Options

A. 55
B. 110
C. 91
D. 182

Answer

To solve the given problem, we need to find the number of quadruplets \((a, b, c, \theta)\) such that the resultant of the three vectors \( \vec{V_1}, \vec{V_2}, \vec{V_3} \) equals \( \lambda \hat{i} \).

### Step-by-Step Solution:

1. **Define the Vectors**:
\[ \vec{V_1} = (\sin \theta) \hat{i} + (\cos \theta) \hat{j} + (a - 3) \hat{k} \]
\[ \vec{V_2} = (\sin \theta + \cos \theta) \hat{i} + (\cos \theta - \sin \theta) \hat{j} + (b - 4) \hat{k} \]
\[ \vec{V_3} = (\cos \theta) \hat{i} + (\sin \theta) \hat{j} + (c - 5) \hat{k} \]

2. **Calculate the Resultant Vector**:
The resultant vector \( \vec{R} = \vec{V_1} + \vec{V_2} + \vec{V_3} \).
- For the \( \hat{i} \) component:
\[ R_i = \sin \theta + (\sin \theta + \cos \theta) + \cos \theta = 2 \sin \theta + 2 \cos \theta \]
- For the \( \hat{j} \) component:
\[ R_j = \cos \theta + (\cos \theta - \sin \theta) + \sin \theta = 2 \cos \theta \]
- For the \( \hat{k} \) component:
\[ R_k = (a - 3) + (b - 4) + (c - 5) = a + b + c - 12 \]

3. **Set the Resultant Equal to \( \lambda \hat{i} \)**:
\[ \vec{R} = \lambda \hat{i} \implies R_j = 0 \text{ and } R_k = 0 \]
This gives us two equations:
- From \( R_j = 0 \):
\[ 2 \cos \theta - 2 \sin \theta = 0 \implies \cos \theta = \sin \theta \implies \tan \theta = 1 \]
- From \( R_k = 0 \):
\[ a + b + c - 12 = 0 \implies a + b + c = 12 \]

4. **Solve for \( \theta \)**:
The equation \( \tan \theta = 1 \) gives:
\[ \theta = \frac{\pi}{4} + n\pi \]
Within the interval \([-\pi, \pi]\), the valid solutions for \( \theta \) are:
\[ \theta = \frac{\pi}{4}, -\frac{3\pi}{4} \]
Thus, there are **2 values** for \( \theta \).

5. **Find the Number of Solutions for \( a, b, c \)**:
We need to find the number of natural number solutions to the equation \( a + b + c = 12 \).
- Using the "stars and bars" theorem, the number of solutions in natural numbers is given by:
\[ \text{Number of solutions} = \binom{n-1}{k-1} = \binom{12-1}{3-1} = \binom{11}{2} = \frac{11 \times 10}{2} = 55 \]

6. **Total Number of Quadruplets**:
The total number of quadruplets \((a, b, c, \theta)\) is:
\[ \text{Total} = \text{Number of solutions for } (a, b, c) \times \text{Number of values for } \theta = 55 \times 2 = 110 \]

### Final Answer:
The number of quadruplets \((a, b, c, \theta)\) is **110**.