Let $$x \in R$$ and let
$$P = \begin{bmatrix}1 & 1 & 1 \\0 & 2 & 2\\0 & 0 & 3\end{bmatrix}, Q = \begin{bmatrix}2 & x & x \\0 & 4 & 0\\x & x & 6\end{bmatrix}$$ and $$R = PQP^{-1}$$
Then which of the following options is/are correct?

There exists a real number x such that PQ = QP

det R = det $$\begin{bmatrix}2 & x & x \\0 & 4 & 0\\x & x & 5\end{bmatrix} + 8,$$ for all $$x \in R$$

For x = 0, if $$R \begin{bmatrix}1 \\a \\b \end{bmatrix} = 6 \begin{bmatrix}1 \\a \\b \end{bmatrix},$$ a + b = 5

For x = 1, there exists a unit vector $$\alpha \widehat{i} + \beta \widehat{j} + \gamma \widehat{k}$$ for which $$R \begin{bmatrix}\alpha \\\beta \\\gamma \end{bmatrix} = \begin{bmatrix}0 \\0 \\0 \end{bmatrix}$$