If `veca=xhati + y hatj + zhatk, vecb= yhati + zhatj + xhatk ` and `vecc= zhati + xhatj+ yhatk `, `" then " vecaxx (vecbxx vecc) `is
(a)parallel to ` ( y- z) hati + (z - x) hatj + (x -y) hatk` (b)orthogonal to `hati + hatj + hatk` (c)orthogonal to ` ( y+z) hati + ( z + x) hatj + ( x + y) hatk` (d)orthogonal to `xhati + yhatj + zhatk`

Options

A. parallel to ` ( y- z) hati + (z - x) hatj + (x -y) hatk`
B. orthogonal to `hati + hatj + hatk`
C. orthogonal to ` ( y+z) hati + ( z + x) hatj + ( x + y) hatk`
D. orthogonal to `xhati + yhatj + zhatk`

Explanation

To solve the problem, we need to find the vector \( \vec{v} = \vec{a} \times (\vec{b} \times \vec{c}) \) and analyze its properties with respect to the given options.

### Step 1: Define the vectors
Given:
\[ \vec{a} = x \hat{i} + y \hat{j} + z \hat{k} \]
\[ \vec{b} = y \hat{i} + z \hat{j} + x \hat{k} \]
\[ \vec{c} = z \hat{i} + x \hat{j} + y \hat{k} \]

### Step 2: Use the vector triple product identity
We can use the identity:
\[ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \]

### Step 3: Calculate \( \vec{a} \cdot \vec{b} \) and \( \vec{a} \cdot \vec{c} \)
First, calculate \( \vec{a} \cdot \vec{b} \):
\[ \vec{a} \cdot \vec{b} = (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (y \hat{i} + z \hat{j} + x \hat{k}) = xy + yz + zx \]

Next, calculate \( \vec{a} \cdot \vec{c} \):
\[ \vec{a} \cdot \vec{c} = (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (z \hat{i} + x \hat{j} + y \hat{k}) = xz + xy + yz \]

### Step 4: Substitute into the vector triple product formula
Now substitute back into the formula:
\[ \vec{v} = (xz + xy + yz) \vec{b} - (xy + yz + zx) \vec{c} \]
Substituting \( \vec{b} \) and \( \vec{c} \):
\[ \vec{v} = (xz + xy + yz)(y \hat{i} + z \hat{j} + x \hat{k}) - (xy + yz + zx)(z \hat{i} + x \hat{j} + y \hat{k}) \]

### Step 5: Expand the expression
Expanding both terms:
\[ \vec{v} = (xz + xy + yz)(y \hat{i}) + (xz + xy + yz)(z \hat{j}) + (xz + xy + yz)(x \hat{k}) - (xy + yz + zx)(z \hat{i}) - (xy + yz + zx)(x \hat{j}) - (xy + yz + zx)(y \hat{k}) \]

### Step 6: Combine like terms
Combine the coefficients of \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \):
\[ \vec{v} = [(xz + xy + yz)y - (xy + yz + zx)z] \hat{i} + [(xz + xy + yz)z - (xy + yz + zx)x] \hat{j} + [(xz + xy + yz)x - (xy + yz + zx)y] \hat{k} \]

### Step 7: Factor out common terms
The expression can be simplified to:
\[ \vec{v} = l \left( (y - z) \hat{i} + (z - x) \hat{j} + (x - y) \hat{k} \right) \]
where \( l = xz + xy + yz \).

### Step 8: Analyze the options
1. **Option (a)**: \( \vec{v} \) is parallel to \( (y - z) \hat{i} + (z - x) \hat{j} + (x - y) \hat{k} \) - **True**.
2. **Option (b)**: Check if \( \vec{v} \) is orthogonal to \( \hat{i} + \hat{j} + \hat{k} \):
\[ \vec{v} \cdot (\hat{i} + \hat{j} + \hat{k}) = (y - z) + (z - x) + (x - y) = 0 \quad \text{(True)} \]
3. **Option (c)**: Check if \( \vec{v} \) is orthogonal to \( (y + z) \hat{i} + (z + x) \hat{j} + (x + y) \hat{k} \):
\[ \vec{v} \cdot [(y + z) \hat{i} + (z + x) \hat{j} + (x + y) \hat{k}] = 0 \quad \text{(True)} \]
4. **Option (d)**: Check if \( \vec{v} \) is orthogonal to \( \vec{a} \):
\[ \vec{v} \cdot \vec{a} = 0 \quad \text{(True)} \]

### Conclusion
All options are correct.