To solve the problem, we need to determine the number of quadruplets \((a, b, c, \theta)\) such that the resultant of the three vectors \(\vec{V_1}\), \(\vec{V_2}\), and \(\vec{V_3}\) is equal to \(\lambda \hat{i}\).
### Step 1: Write down the vectors
The vectors are given as:
\[
\vec{V_1} = (\sin \theta) \hat{i} + (\cos \theta) \hat{j} + (a - 3) \hat{k}
\]
\[
\vec{V_2} = (\sin \theta + \cos \theta) \hat{i} + (\cos \theta - \sin \theta) \hat{j} + (b - 4) \hat{k}
\]
\[
\vec{V_3} = (\cos \theta) \hat{i} + (\sin \theta) \hat{j} + (c - 5) \hat{k}
\]
### Step 2: Find the resultant vector
The resultant vector \(\vec{R}\) is given by:
\[
\vec{R} = \vec{V_1} + \vec{V_2} + \vec{V_3}
\]
Calculating the components:
- **i-component**:
\[
R_i = \sin \theta + (\sin \theta + \cos \theta) + \cos \theta = 2 \sin \theta + \cos \theta
\]
- **j-component**:
\[
R_j = \cos \theta + (\cos \theta - \sin \theta) + \sin \theta = 2 \cos \theta
\]
- **k-component**:
\[
R_k = (a - 3) + (b - 4) + (c - 5) = a + b + c - 12
\]
Thus, the resultant vector can be expressed as:
\[
\vec{R} = (2 \sin \theta + \cos \theta) \hat{i} + (2 \cos \theta) \hat{j} + (a + b + c - 12) \hat{k}
\]
### Step 3: Set the resultant equal to \(\lambda \hat{i}\)
According to the problem, we have:
\[
\vec{R} = \lambda \hat{i}
\]
This implies:
1. \(2 \sin \theta + \cos \theta = \lambda\)
2. \(2 \cos \theta = 0\)
3. \(a + b + c - 12 = 0\)
### Step 4: Solve for \(\theta\)
From the second equation \(2 \cos \theta = 0\):
\[
\cos \theta = 0
\]
This gives us:
\[
\theta = -\frac{\pi}{2} \quad \text{or} \quad \theta = \frac{\pi}{2}
\]
### Step 5: Solve for \(a + b + c\)
From the third equation:
\[
a + b + c = 12
\]
Since \(a, b, c\) are natural numbers, we have \(a, b, c \geq 1\). Thus, we can rewrite:
\[
a' + b' + c' = 9
\]
where \(a' = a - 1\), \(b' = b - 1\), \(c' = c - 1\) (which are non-negative integers).
### Step 6: Count the solutions
The number of non-negative integer solutions to \(a' + b' + c' = 9\) can be found using the stars and bars combinatorial method:
\[
\text{Number of solutions} = \binom{9 + 3 - 1}{3 - 1} = \binom{11}{2} = \frac{11 \times 10}{2} = 55
\]
### Step 7: Consider the values of \(\theta\)
Since \(\theta\) can take 2 values (\(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\)), the total number of quadruplets \((a, b, c, \theta)\) is:
\[
\text{Total quadruplets} = 55 \times 2 = 110
\]
### Conclusion
The number of quadruplets \((a, b, c, \theta)\) is **110**.
