1. The problem asks to evaluate the double integral over the entire plane of the vector function \(\begin{pmatrix}0 \\ y \\ x\end{pmatrix}\) with respect to \(x\) and \(y\) from \(-\infty\) to \(\infty\).\n\n2. The integral is \(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \begin{pmatrix}0 \\ y \\ x\end{pmatrix} \, dx \, dy\). Since integration is linear, we can integrate each component separately.\n\n3. For the first component, \(0\), the integral is \(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} 0 \, dx \, dy = 0\).\n\n4. For the second component, \(y\), the integral is \(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} y \, dx \, dy\). Since \(y\) does not depend on \(x\), we can write \(\int_{-\infty}^{\infty} y \, dx = y \cdot \int_{-\infty}^{\infty} dx\). But \(\int_{-\infty}^{\infty} dx = \infty\), so the integral diverges unless the function is zero or integrable in some other way.\n\n5. Similarly, for the third component, \(x\), the integral is \(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} x \, dx \, dy\). Integrating with respect to \(x\) first, \(\int_{-\infty}^{\infty} x \, dx\) does not converge because the integral of an odd function over symmetric infinite limits is zero only if the integral converges absolutely, but here it is improper and diverges.\n\n6. More precisely, the integral of \(x\) over \(-\infty\) to \(\infty\) is an improper integral that does not converge absolutely, so the double integral does not converge in the usual sense.\n\n7. Therefore, the double integral of the vector function \(\begin{pmatrix}0 \\ y \\ x\end{pmatrix}\) over the entire plane does not converge to a finite vector.\n\n8. If we consider the integral in the sense of the Cauchy principal value, the integrals of \(x\) and \(y\) over symmetric limits are zero because they are odd functions.\n\n9. Hence, the principal value of the integral is \($$\begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}$$\).