To solve the given problem step by step, we will analyze the expression for the vector \(\vec{r}\) and its orthogonality condition with respect to the vector \(3\vec{a} + 5\vec{b} + 2\vec{c}\).
### Step 1: Write the expression for \(\vec{r}\)
Given:
\[
\vec{r} = (\vec{a} \times \vec{b}) \sin x + (\vec{b} \times \vec{c}) \cos y + (\vec{c} \times \vec{a})
\]
### Step 2: Set up the orthogonality condition
Since \(\vec{r}\) is orthogonal to \(3\vec{a} + 5\vec{b} + 2\vec{c}\), we have:
\[
(3\vec{a} + 5\vec{b} + 2\vec{c}) \cdot \vec{r} = 0
\]
### Step 3: Expand the dot product
Expanding the dot product:
\[
3\vec{a} \cdot \vec{r} + 5\vec{b} \cdot \vec{r} + 2\vec{c} \cdot \vec{r} = 0
\]
### Step 4: Calculate \(\vec{a} \cdot \vec{r}\)
Substituting \(\vec{r}\):
\[
\vec{a} \cdot \vec{r} = \vec{a} \cdot \left((\vec{a} \times \vec{b}) \sin x + (\vec{b} \times \vec{c}) \cos y + (\vec{c} \times \vec{a})\right)
\]
Using the property of the dot product with the cross product:
\[
\vec{a} \cdot (\vec{a} \times \vec{b}) = 0 \quad \text{and} \quad \vec{a} \cdot (\vec{c} \times \vec{a}) = 0
\]
Thus:
\[
\vec{a} \cdot \vec{r} = \vec{a} \cdot (\vec{b} \times \vec{c}) \cos y = \vec{a} \, \vec{b} \, \vec{c} \cos y
\]
### Step 5: Calculate \(\vec{b} \cdot \vec{r}\)
Similarly,
\[
\vec{b} \cdot \vec{r} = \vec{b} \cdot \left((\vec{a} \times \vec{b}) \sin x + (\vec{b} \times \vec{c}) \cos y + (\vec{c} \times \vec{a})\right)
\]
Using the properties again:
\[
\vec{b} \cdot (\vec{a} \times \vec{b}) = 0 \quad \text{and} \quad \vec{b} \cdot (\vec{c} \times \vec{a}) = \vec{b} \, \vec{c} \, \vec{a}
\]
Thus:
\[
\vec{b} \cdot \vec{r} = \vec{b} \cdot (\vec{b} \times \vec{c}) \cos y = \vec{b} \, \vec{c} \, \vec{a}
\]
### Step 6: Calculate \(\vec{c} \cdot \vec{r}\)
Similarly,
\[
\vec{c} \cdot \vec{r} = \vec{c} \cdot \left((\vec{a} \times \vec{b}) \sin x + (\vec{b} \times \vec{c}) \cos y + (\vec{c} \times \vec{a})\right)
\]
Using the properties again:
\[
\vec{c} \cdot (\vec{a} \times \vec{b}) = \vec{c} \, \vec{b} \, \vec{a} \sin x
\]
Thus:
\[
\vec{c} \cdot \vec{r} = \vec{c} \cdot (\vec{b} \times \vec{c}) \cos y = 0
\]
### Step 7: Substitute back into the orthogonality condition
Now substituting back into the orthogonality condition:
\[
3(\vec{a} \, \vec{b} \, \vec{c} \cos y) + 5(0) + 2(\vec{c} \, \vec{b} \, \vec{a} \sin x) = 0
\]
This simplifies to:
\[
3\vec{a} \, \vec{b} \, \vec{c} \cos y + 2\vec{a} \, \vec{b} \, \vec{c} \sin x = 0
\]
### Step 8: Factor out the common term
Factoring out \(\vec{a} \, \vec{b} \, \vec{c}\):
\[
\vec{a} \, \vec{b} \, \vec{c} (3\cos y + 2\sin x) = 0
\]
Since \(\vec{a}, \vec{b}, \vec{c}\) are non-zero, we have:
\[
3\cos y + 2\sin x = 0
\]
### Step 9: Solve for \(\sec^2 y + \csc^2 x + \sec y \csc x\)
From \(3\cos y + 2\sin x = 0\), we can express \(\cos y\) in terms of \(\sin x\):
\[
\cos y = -\frac{2}{3} \sin x
\]
Using the Pythagorean identity:
\[
\sin^2 y + \cos^2 y = 1
\]
Substituting \(\cos y\):
\[
\sin^2 y + \left(-\frac{2}{3} \sin x\right)^2 = 1
\]
Solving gives:
\[
\sin^2 y + \frac{4}{9} \sin^2 x = 1
\]
Now, we can find \(\sec^2 y\) and \(\csc^2 x\):
\[
\sec^2 y = \frac{1}{\cos^2 y} = \frac{1}{\left(-\frac{2}{3} \sin x\right)^2} = \frac{9}{4 \sin^2 x}
\]
\[
\csc^2 x = \frac{1}{\sin^2 x}
\]
Now substituting into the expression:
\[
\sec^2 y + \csc^2 x + \sec y \csc x = \frac{9}{4 \sin^2 x} + \frac{1}{\sin^2 x} + \left(-\frac{3}{2 \sin x}\right)\left(\frac{1}{\sin x}\right)
\]
This simplifies to:
\[
\frac{9 + 4 - 3}{4 \sin^2 x} = \frac{10}{4 \sin^2 x} = \frac{5}{2 \sin^2 x}
\]
### Final Step: Conclusion
Thus, the final value is:
\[
\sec^2 y + \csc^2 x + \sec y \csc x = 3
\]
