Let s be the distance between the lines:
`r=2i-j+k+t(2i+j+2k)` (1)
and `r=i+2j+5k+lambda(2i+j+2k)` (2) then `9s^(2)` =__________

Answer

To find the distance \( s \) between the two given lines and then compute \( 9s^2 \), we will follow these steps:

### Step 1: Identify the lines and their parameters
The equations of the lines are given as:
1. \( \mathbf{r} = 2\mathbf{i} - \mathbf{j} + \mathbf{k} + t(2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \)
2. \( \mathbf{r} = \mathbf{i} + 2\mathbf{j} + 5\mathbf{k} + \lambda(2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \)

From these equations, we can identify:
- Direction vector \( \mathbf{b} = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k} \) (same for both lines)
- Point on line 1: \( \mathbf{A_1} = 2\mathbf{i} - \mathbf{j} + \mathbf{k} \)
- Point on line 2: \( \mathbf{A_2} = \mathbf{i} + 2\mathbf{j} + 5\mathbf{k} \)

### Step 2: Calculate the vector \( \mathbf{A_2} - \mathbf{A_1} \)
\[ \mathbf{A_2} - \mathbf{A_1} = (\mathbf{i} + 2\mathbf{j} + 5\mathbf{k}) - (2\mathbf{i} - \mathbf{j} + \mathbf{k}) \]
\[ = (\mathbf{i} - 2\mathbf{i}) + (2\mathbf{j} + \mathbf{j}) + (5\mathbf{k} - \mathbf{k}) \]
\[ = -\mathbf{i} + 3\mathbf{j} + 4\mathbf{k} \]

### Step 3: Compute the cross product \( \mathbf{b} \times (\mathbf{A_2} - \mathbf{A_1}) \)
Using the determinant form:
\[ \mathbf{b} = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}, \quad \mathbf{A_2} - \mathbf{A_1} = \begin{pmatrix} -1 \\ 3 \\ 4 \end{pmatrix} \]
\[ \mathbf{b} \times (\mathbf{A_2} - \mathbf{A_1}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 2 \\ -1 & 3 & 4 \end{vmatrix} \]

Calculating the determinant:
\[ = \mathbf{i} \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 2 \\ -1 & 4 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & 1 \\ -1 & 3 \end{vmatrix} \]
\[ = \mathbf{i} (1 \cdot 4 - 2 \cdot 3) - \mathbf{j} (2 \cdot 4 - 2 \cdot -1) + \mathbf{k} (2 \cdot 3 - 1 \cdot -1) \]
\[ = \mathbf{i} (4 - 6) - \mathbf{j} (8 + 2) + \mathbf{k} (6 + 1) \]
\[ = -2\mathbf{i} - 10\mathbf{j} + 7\mathbf{k} \]

### Step 4: Calculate the magnitude of the cross product
\[ \|\mathbf{b} \times (\mathbf{A_2} - \mathbf{A_1})\| = \sqrt{(-2)^2 + (-10)^2 + 7^2} \]
\[ = \sqrt{4 + 100 + 49} = \sqrt{153} \]

### Step 5: Calculate the magnitude of \( \mathbf{b} \)
\[ \|\mathbf{b}\| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

### Step 6: Compute the distance \( s \)
Using the formula for the distance between two parallel lines:
\[ s = \frac{\|\mathbf{b} \times (\mathbf{A_2} - \mathbf{A_1})\|}{\|\mathbf{b}\|} = \frac{\sqrt{153}}{3} \]

### Step 7: Compute \( 9s^2 \)
\[ s^2 = \left(\frac{\sqrt{153}}{3}\right)^2 = \frac{153}{9} \]
\[ 9s^2 = 9 \cdot \frac{153}{9} = 153 \]

Thus, the final answer is:
\[ \boxed{153} \]