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, we need to determine the minimum value of \( \frac{4}{\pi^2}(x^2 + y^2) \) given that the vector \( \vec{r} \) is orthogonal to \( \vec{a} + \vec{b} + \vec{c} \).

### Step 1: Set up the equation for orthogonality
Since \( \vec{r} \) is orthogonal to \( \vec{a} + \vec{b} + \vec{c} \), we can write:
\[ \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \]

### Step 2: Expand the dot product
Substituting \( \vec{r} \):
\[ \left( (\vec{a} \times \vec{b}) \sin x + (\vec{b} \times \vec{c}) \cos y + 2(\vec{c} \times \vec{a}) \right) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \]

### Step 3: Distribute the dot product
Expanding this gives:
\[ (\vec{a} \times \vec{b}) \sin x \cdot (\vec{a} + \vec{b} + \vec{c}) + (\vec{b} \times \vec{c}) \cos y \cdot (\vec{a} + \vec{b} + \vec{c}) + 2(\vec{c} \times \vec{a}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \]

### Step 4: Analyze each term
1. \( (\vec{a} \times \vec{b}) \cdot \vec{a} = 0 \) (since \( \vec{a} \times \vec{b} \) is perpendicular to \( \vec{a} \))
2. \( (\vec{a} \times \vec{b}) \cdot \vec{b} = 0 \) (same reasoning)
3. \( (\vec{a} \times \vec{b}) \cdot \vec{c} = \vec{a} \cdot \vec{b} \cdot \vec{c} \) (let's denote this as \( V \))

Thus, the first term simplifies to:
\[ V \sin x \]

For the second term:
1. \( (\vec{b} \times \vec{c}) \cdot \vec{a} = \vec{b} \cdot \vec{c} \cdot \vec{a} \) (denote as \( W \))
2. \( (\vec{b} \times \vec{c}) \cdot \vec{b} = 0 \)
3. \( (\vec{b} \times \vec{c}) \cdot \vec{c} = 0 \)

Thus, the second term simplifies to:
\[ W \cos y \]

For the third term:
1. \( (\vec{c} \times \vec{a}) \cdot \vec{a} = 0 \)
2. \( (\vec{c} \times \vec{a}) \cdot \vec{b} = 2 \vec{c} \cdot \vec{a} \cdot \vec{b} \) (denote as \( Z \))
3. \( (\vec{c} \times \vec{a}) \cdot \vec{c} = 0 \)

Thus, the third term simplifies to:
\[ 2Z \]

### Step 5: Combine the results
Putting this all together, we have:
\[ V \sin x + W \cos y + 2Z = 0 \]

### Step 6: Solve for \( \sin x \) and \( \cos y \)
From the equation \( V \sin x + W \cos y + 2Z = 0 \), we can isolate \( \sin x \) and \( \cos y \):
\[ \sin x = -\frac{W \cos y + 2Z}{V} \]

### Step 7: Find the minimum value of \( \frac{4}{\pi^2}(x^2 + y^2) \)
The maximum values of \( \sin x \) and \( \cos y \) are constrained between -1 and 1. Therefore, the minimum occurs when:
\[ \sin x = -1 \quad \text{and} \quad \cos y = -1 \]
This gives:
\[ x = -\frac{\pi}{2}, \quad y = \pi \]

### Step 8: Calculate \( x^2 + y^2 \)
Now we compute:
\[ x^2 + y^2 = \left(-\frac{\pi}{2}\right)^2 + \pi^2 = \frac{\pi^2}{4} + \pi^2 = \frac{\pi^2}{4} + \frac{4\pi^2}{4} = \frac{5\pi^2}{4} \]

### Step 9: Substitute into the expression
Finally, substituting back into the expression:
\[ \frac{4}{\pi^2} \left( x^2 + y^2 \right) = \frac{4}{\pi^2} \cdot \frac{5\pi^2}{4} = 5 \]

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