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# If one root of the determinant $\begin{vmatrix}x&3&7\\2&x&2\\7&6&x\end{vmatrix}=0$ is $-9$ then the other two roots are

$(a)\;2,7\qquad(b)\;2,-7\qquad(c)\;-2,7\qquad(d)\;-2,-7$

Given :
$\begin{vmatrix}x&3&7\\2&x&2\\7&6&x\end{vmatrix}=0$
$\Rightarrow x(x^2-12)-3(2x-14)+7(12-7x)=0$
$\Rightarrow x^3-67x+126=0$
$\Rightarrow (x+9)(x^2-9x+14)=0$
$\Rightarrow (x+9)(x-2)(x-7)=0$
$x=-9,2,7$
Hence the other two roots are 2,7
Hence (a) is the correct answer.