Given the system of linear equations PX = Q with P \in \mathbb{R}^{3 \times 3} and Q \in \mathbb{R}^{3 \times 1}, we need to analyze statements about the LU decomposition where P = LU. Let's evaluate the options one by one:

1. The system PX = Q can be solved by first solving LY = Q and then UX = Y.

   This statement is true. The LU decomposition allows us to express P as the product of a lower triangular matrix L and an upper triangular matrix U. Therefore, solving PX = Q involves two steps:

   • Solve LY = Q for Y using forward substitution, since L is lower triangular.
   • Solve UX = Y for X using backward substitution, since U is upper triangular.
2. If P is invertible, then both L and U are invertible.

   This statement is true. If P is invertible, then none of the diagonal elements of L or U are zero. For lower triangular L and upper triangular U, non-zero diagonal elements imply that both matrices are invertible.

3. If P is singular, then at least one of the diagonal elements of U is zero.

   This statement is true. If P is singular, it does not have an inverse, which implies the determinant of P is zero. In the LU decomposition of P, this zero determinant translates to at least one zero diagonal element in U, rendering U singular.

4. If P is symmetric, then both L and U are symmetric.

   This statement is false. Even if P is symmetric, L and U are generally not symmetric, as L and U have distinct non-diagonal elements which do not adhere to the symmetry of P.

Thus, the correct statements are:

• The system $PX = Q$ can be solved by first solving $LY = Q$ and then $UX = Y$.
• If $P$ is invertible, then both $L$ and $U$ are invertible.
• If $P$ is singular, then at least one of the diagonal elements of $U$ is zero.