Let \(I=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)\) and \(P=\left(\begin{array}{ll}2 & 0 \\ 0 & 3\end{array}\right)\). Let \(Q=\left(\begin{array}{ll}\mathrm{x} & \mathrm{y} \\ \mathrm{z} & 4\end{array}\right)\) for some non-zero real numbers \(x, y\), and \(z\), for which there is \(2 \times 2\) matrix \(R\) with all entries being non-zero real numbers, such that \(Q R=R P\). Then which of the following statements is (are) TRUE?

\(\begin{aligned}
& \mathrm{QR}=\mathrm{RP} \\
& P=\left(\begin{array}{ll}
2 & 0 \\
0 & 3
\end{array}\right) \quad Q=\left(\begin{array}{ll}
x & y \\
z & 4
\end{array}\right) \\
& \left(\begin{array}{ll}
x & y \\
z & 4
\end{array}\right)\left(\begin{array}{ll}
r_1 & r_2 \\
r_3 & r_4
\end{array}\right)=\left(\begin{array}{ll}
r_1 & r_2 \\
r_3 & r_4
\end{array}\right)\left(\begin{array}{ll}
2 & 0 \\
0 & 3
\end{array}\right) \\
& \left.\begin{array}{l}
\mathrm{xr}_1+\mathrm{yr}_3=2 \mathrm{r}_1 \\
\mathrm{zr}_1+4 \mathrm{r}_3=2 \mathrm{r}_3
\end{array}\right\} \rightarrow \frac{\mathrm{x}-2}{-\mathrm{y}}=\frac{\mathrm{r}_3}{\mathrm{r}_1}=\frac{\mathrm{z}}{-2}
\end{aligned}\)
\(2 x-4=y z\)
\(\left.\begin{array}{l}\mathrm{xr}_2+\mathrm{yr}_4=3 \mathrm{r}_2 \\ \mathrm{zr}_2+4 \mathrm{r}_4=3 \mathrm{r}_4\end{array}\right\} \rightarrow \frac{\mathrm{x}-3}{-\mathrm{y}}=\frac{\mathrm{r}_4}{\mathrm{r}_2}=-\mathrm{z} \quad\) \(\mathrm{x}-3=\mathrm{yz}\)
\(\Rightarrow 2 x-4=x-3\) \(\Rightarrow x=1 ~\&~ y z=-2\)
\(\begin{aligned}
& \Rightarrow \mathrm{Q}-\lambda \mathrm{I}=\left(\begin{array}{cc}
\mathrm{x}-\lambda & \mathrm{y} \\
\mathrm{z} & 4-\lambda
\end{array}\right) \\
& |\mathrm{Q}-\lambda \mathrm{I}|=(\lambda-\mathrm{x})(\lambda-4)-\mathrm{yz} \\
& =\lambda^2-(\mathrm{x}+4) \lambda+4 \mathrm{x}-\mathrm{yz} \\
& |\mathrm{Q}-\lambda \mathrm{I}|=\lambda^2-5 \lambda+6
\end{aligned}\)
Now verify the option