If $\mathrm{A}=\left\{\mathrm{x}: {\mathrm{x}}^2-5\mathrm{x}+6=0\right\}, \mathrm{B}=\left\{2, 4\right\}, \mathrm{C}=\left\{4, 5\right\}$, then $\mathrm{A}\times (\mathrm{B}\cap \mathrm{C})$

• A

  $\left\{\left(2, 4\right), \left(3, 4\right)\right\}$

• B

  $\left\{\left(4, 2\right), \left(4, 3\right)\right\}$

• C

  $\left\{\left(2, 4\right), \left(3, 4\right), \left(4, 4\right)\right\}$

• D

  $\left\{\left(2, 2\right), \left(3, 3\right), \left(4, 4\right), \left(5, 5\right)\right\}$