A unit vector coplanar with `veci + vecj + 2veck and veci + 2 vecj + veck` and perpendicular to `veci + vecj + veck ` is _______

Explanation

To find a unit vector that is coplanar with the vectors \( \vec{i} + \vec{j} + 2\vec{k} \) and \( \vec{i} + 2\vec{j} + \vec{k} \), and also perpendicular to the vector \( \vec{i} + \vec{j} + \vec{k} \), we can follow these steps:

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

### Step 2: Find a vector coplanar with \( \vec{a} \) and \( \vec{b} \)
Vectors \( \vec{a} \) and \( \vec{b} \) are coplanar if their scalar triple product is zero:
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = 0 \]

### Step 3: Calculate \( \vec{b} \times \vec{c} \)
First, we need to compute the cross product \( \vec{b} \times \vec{c} \):
\[ \vec{b} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}, \quad \vec{c} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \]
Using the determinant to find the cross product:
\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 1 & 1 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}(2 \cdot 1 - 1 \cdot 1) - \hat{j}(1 \cdot 1 - 1 \cdot 1) + \hat{k}(1 \cdot 1 - 2 \cdot 1) \]
\[ = \hat{i}(2 - 1) - \hat{j}(0) + \hat{k}(1 - 2) \]
\[ = \hat{i} - \hat{k} \]
Thus, \( \vec{b} \times \vec{c} = \hat{i} - \hat{k} \).

### Step 4: Find the dot product with \( \vec{a} \)
Now we need to find \( \vec{a} \cdot (\vec{b} \times \vec{c}) \):
\[ \vec{a} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \]
Calculating the dot product:
\[ \vec{a} \cdot (\hat{i} - \hat{k}) = 1 \cdot 1 + 1 \cdot 0 + 2 \cdot (-1) = 1 + 0 - 2 = -1 \]
Since this is not zero, we need to find a vector \( \vec{d} \) that is perpendicular to \( \vec{c} \) and satisfies the coplanarity condition.

### Step 5: Set up the equations
Let \( \vec{d} = x \hat{i} + y \hat{j} + z \hat{k} \). We have two conditions:
1. \( \vec{d} \cdot (\hat{i} + 2\hat{j} + \hat{k}) = 0 \)
2. \( \vec{d} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0 \)

From the first condition:
\[ x + 2y + z = 0 \quad \text{(Equation 1)} \]
From the second condition:
\[ x + y + z = 0 \quad \text{(Equation 2)} \]

### Step 6: Solve the equations
Subtract Equation 2 from Equation 1:
\[ (x + 2y + z) - (x + y + z) = 0 \implies y = 0 \]
Substituting \( y = 0 \) in Equation 2:
\[ x + 0 + z = 0 \implies z = -x \]
Thus, \( \vec{d} = x \hat{i} + 0 \hat{j} - x \hat{k} = x(\hat{i} - \hat{k}) \).

### Step 7: Find the unit vector
To make \( \vec{d} \) a unit vector:
\[ \sqrt{x^2 + 0^2 + (-x)^2} = 1 \implies \sqrt{2x^2} = 1 \implies x = \pm \frac{1}{\sqrt{2}} \]
Thus, the unit vector is:
\[ \vec{d} = \pm \frac{1}{\sqrt{2}}(\hat{i} - \hat{k}) \]

### Final Answer
The required unit vector is:
\[ \frac{1}{\sqrt{2}}(\hat{i} - \hat{k}) \text{ or } -\frac{1}{\sqrt{2}}(\hat{i} - \hat{k}) \]