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# If 5 is one root of the equation $\begin{vmatrix}x&3&7\\2&x&-2\\7&8&x\end{vmatrix}=0$ then the other two roots of the equation are

$\begin{array}{1 1}(A)\;-2,7&(B)\;-2,-7\\(C)\;2,7&(D)\;2,-7\end{array}$

Given
$\begin{vmatrix}x&3&7\\2&x&-2\\7&8&x\end{vmatrix}=0$
On expanding along $R_1$ we get
$x(x^2+16)-3(2x+14)+7(16-7x)=0$
$\Rightarrow x(x^2+16)-3(2x+14)+7(16-7x)=0$
$\Rightarrow x^3-39x+70=0$
$\Rightarrow (x-5)(x^2+5x-14)=0$
$x=5$ is given
$(x-5)(x+7)(x-2)=0$
$x=-7,2$
Hence (D) is the correct answer.