Find `lambda + mu` if `(2hat(i) + 6hat(j) + 27 hat(k)) xx (hat(i) + lambda hat(j) + mu hat(k)) = vec(0)`.

Answer

To solve the problem, we need to find the values of \(\lambda\) and \(\mu\) such that the cross product of the vectors \( \mathbf{A} = 2\hat{i} + 6\hat{j} + 27\hat{k} \) and \( \mathbf{B} = \hat{i} + \lambda\hat{j} + \mu\hat{k} \) equals the zero vector.

### Step-by-Step Solution:

1. **Set Up the Cross Product**:
We need to compute the cross product \( \mathbf{A} \times \mathbf{B} \):
\[ \mathbf{A} \times \mathbf{B} = (2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k}) \]

2. **Use the Determinant for Cross Product**:
We can express the cross product using a determinant:
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{vmatrix} \]

3. **Calculate the Determinant**:
Expanding the determinant, we have:
\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} 6 & 27 \\ \lambda & \mu \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 27 \\ 1 & \mu \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 6 \\ 1 & \lambda \end{vmatrix} \]

Calculating each of these 2x2 determinants:
- For \(\hat{i}\):
\[ 6\mu - 27\lambda \]
- For \(\hat{j}\):
\[ 2\mu - 27 \]
- For \(\hat{k}\):
\[ 2\lambda - 6 \]

Thus, we have:
\[ \mathbf{A} \times \mathbf{B} = (6\mu - 27\lambda)\hat{i} - (2\mu - 27)\hat{j} + (2\lambda - 6)\hat{k} \]

4. **Set the Cross Product to Zero**:
Since \( \mathbf{A} \times \mathbf{B} = \vec{0} \), we can set each component to zero:
\[ 6\mu - 27\lambda = 0 \quad (1) \]
\[ 2\mu - 27 = 0 \quad (2) \]
\[ 2\lambda - 6 = 0 \quad (3) \]

5. **Solve the Equations**:
From equation (2):
\[ 2\mu = 27 \implies \mu = \frac{27}{2} \]

From equation (3):
\[ 2\lambda = 6 \implies \lambda = 3 \]

6. **Find \(\lambda + \mu\)**:
Now we can find \(\lambda + \mu\):
\[ \lambda + \mu = 3 + \frac{27}{2} \]
To add these, convert \(3\) into a fraction:
\[ 3 = \frac{6}{2} \]
Thus,
\[ \lambda + \mu = \frac{6}{2} + \frac{27}{2} = \frac{33}{2} \]

### Final Answer:
\[ \lambda + \mu = \frac{33}{2} \]