If $P$ and $Q$ are two non-zero square matrices of the same order such that the product $PQ = 0$,then ........

The value of $\left| {\begin{array}{*{20}{c}}1&{\cos (\beta - \alpha )}&{\cos (\gamma - \alpha )}\\{\cos (\alpha - \beta )}&1&{\cos (\gamma - \beta )}\\{\cos (\alpha - \gamma )}&{\cos (\beta - \gamma )}&1\end{array}} \right|$ is

For $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & 3 \\ 1 & 1 & 0 \end{bmatrix}$,if $A^3 - 2A^2 + kA - 4I_3 = 0$,then $k = $ . . . . . . .

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?
$(1)$ For $x = 1$,there exists a unit vector $\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ for which $R \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$.
$(2)$ There exists a real number $x$ such that $PQ = QP$.
$(3)$ $\det R = \det \begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 5 \end{bmatrix} + 8$,for all $x \in R$.
$(4)$ For $x = 0$,if $R \begin{bmatrix} 1 \\ a \\ b \end{bmatrix} = 6 \begin{bmatrix} 1 \\ a \\ b \end{bmatrix}$,then $a + b = 5$.

If $A = \begin{bmatrix} 1 & a & 3 \\ b & 2 & c \\ 3 & d & 4 \end{bmatrix}$ is a symmetric matrix and $B = \begin{bmatrix} 0 & 5 & b \\ -5 & 0 & -7 \\ 6 & c & 0 \end{bmatrix}$ is a skew-symmetric matrix, then $AB = $

Let $A$ and $B$ be orthogonal matrices and $\operatorname{det}(A) + \operatorname{det}(B) = 0$. Then