Let `vecr=(veca xx vecb)sinx+(vecb xx vec c)cosy+2(vec c xx vec a)`, where `veca, vecb, vec c` are non-zero and non-coplanar vectors. If `vecr` is orthogonal to `veca+vecb+vec c`, then find the minimum value of `(4)/(pi^(2))(x^(2)+y^(2))`.

Answer

To solve the problem step by step, we start with the given vector expression and the condition of orthogonality.

### Step 1: Write the given vector expression
We have:
\[ \vec{r} = (\vec{a} \times \vec{b}) \sin x + (\vec{b} \times \vec{c}) \cos y + 2(\vec{c} \times \vec{a}) \]

### Step 2: Use the orthogonality condition
Since \(\vec{r}\) is orthogonal to \(\vec{a} + \vec{b} + \vec{c}\), we have:
\[ \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \]
This expands to:
\[ \vec{r} \cdot \vec{a} + \vec{r} \cdot \vec{b} + \vec{r} \cdot \vec{c} = 0 \]

### Step 3: Calculate \(\vec{r} \cdot \vec{a}\)
Using the properties of the dot product and the cross product:
\[ \vec{r} \cdot \vec{a} = \left((\vec{a} \times \vec{b}) \sin x\right) \cdot \vec{a} + \left((\vec{b} \times \vec{c}) \cos y\right) \cdot \vec{a} + \left(2(\vec{c} \times \vec{a})\right) \cdot \vec{a} \]
The first term is zero because \(\vec{a} \times \vec{b} \cdot \vec{a} = 0\). The third term is also zero because \(\vec{c} \times \vec{a} \cdot \vec{a} = 0\). Thus:
\[ \vec{r} \cdot \vec{a} = (\vec{b} \times \vec{c}) \cdot \vec{a} \cos y \]

### Step 4: Calculate \(\vec{r} \cdot \vec{b}\)
Similarly,
\[ \vec{r} \cdot \vec{b} = \left((\vec{a} \times \vec{b}) \sin x\right) \cdot \vec{b} + \left((\vec{b} \times \vec{c}) \cos y\right) \cdot \vec{b} + \left(2(\vec{c} \times \vec{a})\right) \cdot \vec{b} \]
Again, the first and second terms are zero. Thus:
\[ \vec{r} \cdot \vec{b} = 2(\vec{c} \times \vec{a}) \cdot \vec{b} \]

### Step 5: Calculate \(\vec{r} \cdot \vec{c}\)
Following the same logic:
\[ \vec{r} \cdot \vec{c} = \left((\vec{a} \times \vec{b}) \sin x\right) \cdot \vec{c} + \left((\vec{b} \times \vec{c}) \cos y\right) \cdot \vec{c} + \left(2(\vec{c} \times \vec{a})\right) \cdot \vec{c} \]
The second and third terms are zero, so:
\[ \vec{r} \cdot \vec{c} = (\vec{a} \times \vec{b}) \cdot \vec{c} \sin x \]

### Step 6: Combine the results
Now we substitute back into the orthogonality condition:
\[ (\vec{b} \times \vec{c}) \cdot \vec{a} \cos y + 2(\vec{c} \times \vec{a}) \cdot \vec{b} + (\vec{a} \times \vec{b}) \cdot \vec{c} \sin x = 0 \]

### Step 7: Factor out the scalar triple product
Let \(V = \vec{a} \cdot (\vec{b} \times \vec{c})\) (the scalar triple product):
\[ V \cos y + 2V + V \sin x = 0 \]
This simplifies to:
\[ V(\cos y + 2 + \sin x) = 0 \]

### Step 8: Solve for \(\cos y + \sin x\)
Since \(V \neq 0\) (as \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar), we have:
\[ \cos y + \sin x + 2 = 0 \implies \cos y + \sin x = -2 \]
This is not possible since both \(\cos y\) and \(\sin x\) are bounded between -1 and 1.

### Step 9: Find minimum value
To minimize \( \frac{4}{\pi^2}(x^2 + y^2) \), we note that:
\[ \sin x = -1 \quad \text{and} \quad \cos y = -1 \]
This gives:
\[ x = -\frac{\pi}{2}, \quad y = \pi \]
Thus:
\[ x^2 + y^2 = \left(-\frac{\pi}{2}\right)^2 + \pi^2 = \frac{\pi^2}{4} + \pi^2 = \frac{5\pi^2}{4} \]

### Step 10: Calculate the final value
Now substituting back:
\[ \frac{4}{\pi^2} \left(\frac{5\pi^2}{4}\right) = 5 \]

### Final Answer
The minimum value of \( \frac{4}{\pi^2}(x^2 + y^2) \) is:
\[ \boxed{5} \]