Let `veca=hati+2hatj+3hatk`, `vecb=2hati+3hatj+hatk, vecc=hatk+hati and (vecx xx vecb)=(veca xx vecc)xxvecb`. If `vecx .veca=0` , then `|vecx|` is equal to use `sqrt3=1.73`)``

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the vectors
We are given:
- \(\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}\)
- \(\vec{b} = 2\hat{i} + 3\hat{j} + \hat{k}\)
- \(\vec{c} = \hat{k} + \hat{i}\)

### Step 2: Express \(\vec{x}\)
Let \(\vec{x} = x_x \hat{i} + x_y \hat{j} + x_z \hat{k}\).

### Step 3: Use the dot product condition
We know that \(\vec{x} \cdot \vec{a} = 0\). Therefore,
\[ \vec{x} \cdot \vec{a} = (x_x \hat{i} + x_y \hat{j} + x_z \hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) = x_x + 2x_y + 3x_z = 0 \]
This gives us our first equation:
\[ x_x + 2x_y + 3x_z = 0 \quad \text{(Equation 1)} \]

### Step 4: Use the cross product condition
We are given that \(\vec{x} \times \vec{b} = (\vec{a} \times \vec{c}) \times \vec{b}\).

First, we need to find \(\vec{a} \times \vec{c}\):
\[ \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 1 & 0 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}(2 \cdot 1 - 3 \cdot 0) - \hat{j}(1 \cdot 1 - 3 \cdot 1) + \hat{k}(1 \cdot 0 - 2 \cdot 1) \]
\[ = 2\hat{i} - (1 - 3)\hat{j} - 2\hat{k} = 2\hat{i} + 2\hat{j} - 2\hat{k} \]
Thus, \(\vec{a} \times \vec{c} = 2\hat{i} + 2\hat{j} - 2\hat{k}\).

### Step 5: Substitute into the cross product equation
Now, substituting into the equation:
\[ \vec{x} \times \vec{b} = (2\hat{i} + 2\hat{j} - 2\hat{k}) \times \vec{b} \]
Let’s calculate \((2\hat{i} + 2\hat{j} - 2\hat{k}) \times (2\hat{i} + 3\hat{j} + \hat{k})\):
\[ = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & -2 \\ 2 & 3 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}(2 \cdot 1 - (-2) \cdot 3) - \hat{j}(2 \cdot 1 - (-2) \cdot 2) + \hat{k}(2 \cdot 3 - 2 \cdot 2) \]
\[ = \hat{i}(2 + 6) - \hat{j}(2 + 4) + \hat{k}(6 - 4) \]
\[ = 8\hat{i} - 6\hat{j} + 2\hat{k} \]

### Step 6: Set up the equation
Now we have:
\[ \vec{x} \times \vec{b} = 8\hat{i} - 6\hat{j} + 2\hat{k} \]
Since \(\vec{x} \times \vec{b} = \lambda \vec{b}\) for some scalar \(\lambda\), we can express this as:
\[ \vec{x} - (2\hat{i} + 2\hat{j} - 2\hat{k}) = \lambda (2\hat{i} + 3\hat{j} + \hat{k}) \]

### Step 7: Solve for \(\vec{x}\)
From the previous equations, we can express \(\vec{x}\) in terms of \(\lambda\):
\[ \vec{x} = \lambda(2\hat{i} + 3\hat{j} + \hat{k}) + (2\hat{i} + 2\hat{j} - 2\hat{k}) \]

### Step 8: Find the magnitude of \(\vec{x}\)
To find \(|\vec{x}|\):
\[ |\vec{x}| = \sqrt{(2 + 2\lambda)^2 + (2 + 3\lambda)^2 + (-2 + \lambda)^2} \]

### Step 9: Substitute values and solve
Substituting \(\lambda = 0\) (as derived from previous steps):
\[ |\vec{x}| = \sqrt{(2)^2 + (2)^2 + (-2)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3} \]

### Final Step: Use the value of \(\sqrt{3}\)
Using \(\sqrt{3} \approx 1.73\):
\[ |\vec{x}| \approx 2 \times 1.73 = 3.46 \]

Thus, the magnitude \(|\vec{x}|\) is approximately \(3.46\).