If `vecA,vecB,vecC` are non-coplanar vectors than `( vecA . vecB xx vecC )/(vecCxxvecA.vecB)+(vecB. vecA xx vecC)/( vecC. vecA xx vecB)` is equal to

Answer

To solve the problem, we need to simplify the given expression step by step. The expression is:

\[ \frac{\vec{A} \cdot (\vec{B} \times \vec{C})}{\vec{C} \cdot (\vec{A} \times \vec{B})} + \frac{\vec{B} \cdot (\vec{A} \times \vec{C})}{\vec{C} \cdot (\vec{A} \times \vec{B})} \]

### Step 1: Identify the Scalar Triple Products
The terms in the numerators can be recognized as scalar triple products. We can rewrite the expression as:

\[ \frac{[\vec{A}, \vec{B}, \vec{C}]}{[\vec{C}, \vec{A}, \vec{B}]} + \frac{[\vec{B}, \vec{A}, \vec{C}]}{[\vec{C}, \vec{A}, \vec{B}]} \]

Where \([\vec{X}, \vec{Y}, \vec{Z}]\) denotes the scalar triple product of vectors \(\vec{X}, \vec{Y}, \vec{Z}\).

### Step 2: Combine the Fractions
Since both terms have the same denominator, we can combine them:

\[ \frac{[\vec{A}, \vec{B}, \vec{C}] + [\vec{B}, \vec{A}, \vec{C}]}{[\vec{C}, \vec{A}, \vec{B}]} \]

### Step 3: Simplify the Numerator
Now, we need to simplify the numerator. The scalar triple product has the property that swapping two vectors changes the sign:

\[ [\vec{B}, \vec{A}, \vec{C}] = -[\vec{A}, \vec{B}, \vec{C}] \]

Thus, we can write:

\[ [\vec{A}, \vec{B}, \vec{C}] + [\vec{B}, \vec{A}, \vec{C}] = [\vec{A}, \vec{B}, \vec{C}] - [\vec{A}, \vec{B}, \vec{C}] = 0 \]

### Step 4: Final Expression
Substituting back into our combined fraction gives:

\[ \frac{0}{[\vec{C}, \vec{A}, \vec{B}]} = 0 \]

Thus, the value of the entire expression is:

\[ \boxed{0} \]