If $x = - 9$ is a root of $\left| {\begin{array}{llllllllllllllllllll}x&3&7\\2&x&2\\7&6&x\end{array}} \right| = 0$, then other two roots are…………….

Since $\left| {\begin{array}{llllllllllllllllllll}x&3&7\\2&x&2\\7&6&x\end{array}} \right| = 0$

Expanding along ${R_1}$,

$x\left( {{x^2} - 12} \right) - 3(2x - 14) + 7(12 - 7x) = 0$

$\Rightarrow$ ${x^3} - 12x - 6x + 42 + 84 - 49x = 0$
$\Rightarrow$ ${x^3} - 67x + 126 = 0$

……….(i)
Here, $126 \times 1 = 9 \times 2 \times 7$

For $x = 2,$ ${2^3} - 67 \times 2 + 126 = 134 - 134 = 0$

Hence, $x = 2$ is a root.
For $x = 7,$ ${7^3} - 67 \times 7 + 126 = 469 - 469 = 0$

Hence, $x = 7$ is also a root.