Let x ∈ R and let P = [ cases 1 & 1 & 1 0 & 2 & 2 0 & 0 & 3 cases...

$$\begin{array}{ll}P = 1 1 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3\end{array}$$Q = \left[$\begin{cases}2 & x & x \\ 0 & 4 & 0 \\ x & x & 6\end{cases}$\right]$<br />Option$( C )$<br />Now$R = PQ{P}^{- 1}$<br />$det \left(R\right) = det \left(R\right)det \left(Q\right)det \left({P}^{- 1}\right)$<br />$\begin{cases}det \left(R\right) = det \left(Q\right) & \left(det \left({P}^{- 1}\right) = \dfrac{1}{det \left(P\right)}\right)\end{cases}$<br />$det \left(R\right) = det \left[$\begin{cases}2 & x & x \\ 0 & 4 & 0 \\ x & x & 6\end{cases}$\right]$<br />$det \left(R\right) = 48 - 4{x}^{2}$<br />now$det \left[$\begin{cases}2 & x & x \\ 0 & 4 & 0 \\ x & x & 5\end{cases}$\right] = 40 - 4{x}^{2}$<br />$det R = det \left[$\begin{cases}2 & x & x \\ 0 & 4 & 0 \\ x & x & 5\end{cases}$\right] + 8$<br />Option \left(A\right)$
R\left[$\begin{array}{ll}\alpha \\ \beta \\ \gamma\end{array}$\right] = \left[$\begin{array}{ll}0 \\ 0 \\ 0\end{array}$\right] must have not trivial solution
So $det \left(R\right) = 0$
$48 - 4{x}^{2} = 0 ⇒ x = \pm 2\sqrt{3}$
Option \left(D\right)$<br />$R\left[$$\begin{array}{ll}1 \\ a \\ b\end{array}$$\right] = \left[$$\begin{array}{ll}6 \\ 6a \\ 6b\end{array}$$\right]$<br />$PQ{P}^{- 1}\left[$$\begin{array}{ll}1 \\ a \\ b\end{array}$$\right] = \left[$$\begin{array}{ll}6 \\ 6a \\ 6b\end{array}$$\right]&hellip;(i)<br />{P}^{- 1} = \dfrac{1}{6}\left[$\begin{cases}6 & - 3 & 0 \\ 0 & 3 & - 2 \\ 0 & 0 & 2\end{cases}$\right]$<br />Putting$P, Q, {P}^{- 1}$in equation (i)<br />$\dfrac{1}{6}\left[$\begin{cases}12 & 6 & 4 \\ 0 & 24 & 8 \\ 0 & 0 & 36\end{cases}$\right]\left[$$\begin{array}{ll}1 \\ a \\ b\end{array}$$\right] = \left[$$\begin{array}{ll}6 \\ 6a \\ 6b\end{array}$$\right]$<br />$⇒ 12 + 6a + 4b = 36$<br />$24a + 8b = 36a$<br />$a = 2$and$b = 3$<br />$a + b = 5$<br />Option \left(B\right)$
$PQ = QP$
$PQ = QP ⇒ PQ{P}^{- 1} = Q$
$R = Q$
Not possible for any value of $x .