Let the given matrix be $A = \left[ {\begin{matrix} 2 & 1 & 3 \ 1 & 0 & 2 \ 3 & 2 & 1 \ \end{matrix} } \right]$. We can observe that A is a symmetric matrix, i.e., $A^T = A$. The matrices $P_k$ are the six 3x3 permutation matrices. A key property of permutation matrices is that they are orthogonal, meaning $P_k P_k^T = P_k^T P_k = I$, where $I$ is the identity matrix.

The matrix X is defined as $X = \sum\limits_{k = 1}^6 {{P_k}} A P_k^T$

We will evaluate each option step-by-step.

Option A: X is a symmetric matrix

1. To check if X is symmetric, we need to compute its transpose, $X^T$. $X^T = {\left( {\sum\limits_{k = 1}^6 {{P_k}A P_k^T} } \right)^T}$
2. Using the property $(A+B)^T = A^T+B^T$ and $(ABC)^T = C^T B^T A^T$, we get: X^T = \sum\limits_{k = 1}^6 {{{\left( {{P_k}A P_k^T} \right)}^T}} = \sum\limits_{k = 1}^6 {{{\left( {P_k^T} \right)}^T}{A^T}P_k^T}}
3. We know that $(P_k^T)^T = P_k$ and the given matrix A is symmetric ($A^T = A$). Substituting these into the expression for $X^T$: $X^T = \sum\limits_{k = 1}^6 {{P_k}A P_k^T} = X$
4. Since $X^T = X$, the matrix X is symmetric. Thus, option A is correct.

Option B: The sum of diagonal entries of X is 18

1. The sum of diagonal entries of a matrix is its trace, denoted by Tr(). We need to calculate Tr(X). $Tr(X) = Tr\left( {\sum\limits_{k = 1}^6 {{P_k}A P_k^T} } \right) = \sum\limits_{k = 1}^6 {Tr({P_k}A P_k^T)}$
2. Using the cyclic property of the trace, $Tr(ABC) = Tr(BCA)$, we have: $Tr({P_k}A P_k^T) = Tr(A P_k^T {P_k})$
3. Since $P_k$ are permutation matrices, they are orthogonal, so $P_k^T P_k = I$. $Tr(A P_k^T {P_k}) = Tr(AI) = Tr(A)$
4. The trace of A is the sum of its diagonal elements: $Tr(A) = 2 + 0 + 1 = 3$.
5. Now we can find the trace of X: $Tr(X) = \sum\limits_{k = 1}^6 {Tr(A)} = 6 \times Tr(A) = 6 \times 3 = 18$
6. The sum of diagonal entries of X is 18. Thus, option B is correct.

Option D: If $X\left[ {\begin{matrix} 1 \ 1 \ 1 \ \end{matrix} } \right] = \alpha \left[ {\begin{matrix} 1 \ 1 \ 1 \ \end{matrix} } \right]$, then $\alpha = 30$

1. Let $v = \left[ {\begin{matrix} 1 \ 1 \ 1 \ \end{matrix} } \right]$. The equation is $Xv = \alpha v$. This means $v$ is an eigenvector of X with eigenvalue $\alpha$. $Xv = \left( {\sum\limits_{k = 1}^6 {{P_k}A P_k^T} } \right)v = \sum\limits_{k = 1}^6 {{P_k}A P_k^T v}$
2. A permutation matrix $P_k$ acting on a vector permutes its components. For the vector $v$, all components are 1, so any permutation of its components results in the same vector $v$. Therefore, $P_k^T v = v$ for all $k$. $Xv = \sum\limits_{k = 1}^6 {{P_k}A v}$
3. First, let's calculate $Av$: $Av = \left[ {\begin{matrix} 2 & 1 & 3 \ 1 & 0 & 2 \ 3 & 2 & 1 \ \end{matrix} } \right]\left[ {\begin{matrix} 1 \ 1 \ 1 \ \end{matrix} } \right] = \left[ {\begin{matrix} {2 + 1 + 3} \ {1 + 0 + 2} \ {3 + 2 + 1} \ \end{matrix} } \right] = \left[ {\begin{matrix} 6 \ 3 \ 6 \ \end{matrix} } \right]$
4. Let $w = Av = \left[ {\begin{matrix} 6 \ 3 \ 6 \ \end{matrix} } \right]$. Then $Xv = \sum\limits_{k = 1}^6 {{P_k}w}$.
5. The sum of all six 3x3 permutation matrices results in a matrix where every entry is 2. This is because for any position $(i,j)$, there are $(3-1)! = 2$ permutations that map row $i$ to row $j$. So, $\sum\limits_{k = 1}^6 P_k = \left[ {\begin{matrix} 2 & 2 & 2 \ 2 & 2 & 2 \ 2 & 2 & 2 \ \end{matrix} } \right] = 2J$, where J is the matrix of all ones.
6. Now we compute $Xv$: $Xv = \left( {\sum\limits_{k = 1}^6 {{P_k}} } \right)w = (2J)w = 2\left[ {\begin{matrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \ \end{matrix} } \right]\left[ {\begin{matrix} 6 \ 3 \ 6 \ \end{matrix} } \right] = 2\left[ {\begin{matrix} {6 + 3 + 6} \ {6 + 3 + 6} \ {6 + 3 + 6} \ \end{matrix} } \right] = 2\left[ {\begin{matrix} {15} \ {15} \ {15} \ \end{matrix} } \right] = \left[ {\begin{matrix} {30} \ {30} \ {30} \ \end{matrix} } \right] = 30\left[ {\begin{matrix} 1 \ 1 \ 1 \ \end{matrix} } \right] = 30v$
7. Comparing $Xv = 30v$ with $Xv = \alpha v$, we get $\alpha = 30$. Thus, option D is correct.

Option C: X $-$ 30I is an invertible matrix

1. From the analysis of option D, we found that $Xv = 30v$ for the non-zero vector $v = \left[ {\begin{matrix} 1 \ 1 \ 1 \ \end{matrix} } \right]$.
2. This equation can be rewritten as $Xv - 30v = 0$, or $(X - 30I)v = 0$.
3. This shows that 30 is an eigenvalue of the matrix X, and $v$ is a corresponding eigenvector.
4. A matrix $M$ is invertible if and only if its determinant is non-zero. The determinant is the product of its eigenvalues. Since 30 is an eigenvalue of X, one of the eigenvalues of the matrix $(X - 30I)$ is $30 - 30 = 0$.
5. Since one of the eigenvalues of $(X - 30I)$ is 0, its determinant is 0. A matrix with a determinant of 0 is singular (not invertible).
6. Therefore, $X - 30I$ is not an invertible matrix. Thus, option C is incorrect.

Final Conclusion: Options A, B, and D are correct.