Let `veca=hati-hatj,b=hati+2hatj+2hatk,vecc=hati-hatj+hatk and vecd = -2hati-hatj-hatk` then the shortest distance between the lines `vecr=veca+tvecb and vecr=vecc+pvecd` is k , then the value of `1/k^2` is

Answer

To find the shortest distance \( k \) between the two lines given by the vectors \( \vec{r} = \vec{a} + t\vec{b} \) and \( \vec{r} = \vec{c} + p\vec{d} \), we will follow these steps:

### Step 1: Define the vectors
Given:
- \( \vec{a} = \hat{i} - \hat{j} \)
- \( \vec{b} = \hat{i} + 2\hat{j} + 2\hat{k} \)
- \( \vec{c} = \hat{i} - \hat{j} + \hat{k} \)
- \( \vec{d} = -2\hat{i} - \hat{j} - \hat{k} \)

### Step 2: Calculate \( \vec{c} - \vec{a} \)
\[ \vec{c} - \vec{a} = (\hat{i} - \hat{j} + \hat{k}) - (\hat{i} - \hat{j}) = \hat{k} \]

### Step 3: Calculate \( \vec{b} \times \vec{d} \)
To find \( \vec{b} \times \vec{d} \), we set up the determinant:
\[ \vec{b} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}, \quad \vec{d} = \begin{pmatrix} -2 \\ -1 \\ -1 \end{pmatrix} \]
The cross product is given by:
\[ \vec{b} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ -2 & -1 & -1 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 2 & 2 \\ -1 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ -2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ -2 & -1 \end{vmatrix} \]
\[ = \hat{i} (2 \cdot (-1) - 2 \cdot (-1)) - \hat{j} (1 \cdot (-1) - 2 \cdot (-2)) + \hat{k} (1 \cdot (-1) - 2 \cdot (-2)) \]
\[ = \hat{i} ( -2 + 2) - \hat{j} (-1 + 4) + \hat{k} (-1 + 4) \]
\[ = 0\hat{i} + 3\hat{j} + 3\hat{k} = 3\hat{j} + 3\hat{k} \]

### Step 4: Calculate the modulus of \( \vec{b} \times \vec{d} \)
\[ |\vec{b} \times \vec{d}| = \sqrt{0^2 + 3^2 + 3^2} = \sqrt{0 + 9 + 9} = \sqrt{18} = 3\sqrt{2} \]

### Step 5: Calculate the shortest distance \( k \)
Using the formula for the shortest distance between two lines:
\[ d = \frac{|(\vec{c} - \vec{a}) \cdot (\vec{b} \times \vec{d})|}{|\vec{b} \times \vec{d}|} \]
Substituting the values:
\[ d = \frac{|\hat{k} \cdot (3\hat{j} + 3\hat{k})|}{3\sqrt{2}} \]
Calculating the dot product:
\[ \hat{k} \cdot (3\hat{j} + 3\hat{k}) = 0 + 3 = 3 \]
Thus,
\[ d = \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Step 6: Find \( \frac{1}{k^2} \)
Since \( k = \frac{1}{\sqrt{2}} \):
\[ k^2 = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
Thus,
\[ \frac{1}{k^2} = 2 \]

### Final Answer
The value of \( \frac{1}{k^2} \) is \( \boxed{2} \).