If one of the roots of left|begin{array}{lll} | vedclass

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(A) Given the determinant equation: $\left|\begin{array}{lll}3 & 5 & x \ 7 & x & 7 \ x & 5 & 3\end{array}ight|=0$ Expanding along the first row: $

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$3, 7$

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(A) Given the determinant equation: $\left|\begin{array}{lll}3 & 5 & x \ 7 & x & 7 \ x & 5 & 3\end{array}ight|=0$ Expanding along the first row: $3(3x - 35) - 5(21 - 7x) + x(35 - x^2) = 0$ $9x - 105 - 105 + 35x + 35x - x^3 = 0$ $-x^3 + 79x - 210 = 0$ $x^3 - 79x + 210 = 0$ Since $x = -10$ is a root,$(x + 10)$ is a factor. Dividing $x^3 - 79x + 210$ by $(x + 10)$ gives $(x^2 - 10x + 21)$. Factoring the quadratic: $(x^2 - 10x + 21) = (x - 3)(x - 7)$. Thus,the equation is $(x + 10)(x - 3)(x - 7) = 0$. The roots are $x = -10, 3, 7$. Therefore,the other roots are $3$ and $7$.

If one of the roots of $\left|\begin{array}{lll}3 & 5 & x \\ 7 & x & 7 \\ x & 5 & 3\end{array}\right|=0$ is $-10$,then the other roots are

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