Let `vec(a),vec(b),vec( c )` be three non-coplanar vectors.
If `(x+2y+z)(vec(b)xx vec( c ))+(x-y)(vec( c )xx vec(a))+(x+y+2)(vec(a)xx vec(b))=vec(0)` for some `x, y, z in R`, then `x^(2)-y^(2)+z^(2)` is equal to _____________

Answer

To solve the problem, we start with the equation given in the question:

\[ (x + 2y + z)(\vec{b} \times \vec{c}) + (x - y)(\vec{c} \times \vec{a}) + (x + y + 2)(\vec{a} \times \vec{b}) = \vec{0} \]

Since \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar vectors, the only way for this equation to hold true is if the coefficients of each vector product are equal to zero. This gives us three equations to solve:

1. \(x + 2y + z = 0\) (Equation 1)
2. \(x - y = 0\) (Equation 2)
3. \(x + y + 2 = 0\) (Equation 3)

### Step 1: Solve Equation 2 for \(y\)

From Equation 2:

\[ x - y = 0 \implies y = x \]

### Step 2: Substitute \(y\) into Equation 3

Now substitute \(y = x\) into Equation 3:

\[ x + x + 2 = 0 \implies 2x + 2 = 0 \]

### Step 3: Solve for \(x\)

Rearranging gives:

\[ 2x = -2 \implies x = -1 \]

### Step 4: Find \(y\) using \(x\)

Using the value of \(x\) in Equation 2:

\[ y = x = -1 \]

### Step 5: Substitute \(x\) and \(y\) into Equation 1 to find \(z\)

Now substitute \(x\) and \(y\) into Equation 1:

\[ -1 + 2(-1) + z = 0 \]

This simplifies to:

\[ -1 - 2 + z = 0 \implies z - 3 = 0 \implies z = 3 \]

### Step 6: Calculate \(x^2 - y^2 + z^2\)

Now we have:

- \(x = -1\)
- \(y = -1\)
- \(z = 3\)

We need to find \(x^2 - y^2 + z^2\):

\[ x^2 - y^2 + z^2 = (-1)^2 - (-1)^2 + (3)^2 \]

Calculating each term:

\[ = 1 - 1 + 9 = 0 + 9 = 9 \]

Thus, the final answer is:

\[ \boxed{9} \]