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(D) Given the inequality $a^2+b^2+c^2-ab-bc-ca \leq 0$. Multiplying by $2$ and rearranging,we get $(a-b)^2 + (b-c)^2 + (c-a)^2 \leq 0$. Since the sum

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$24$

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(D) Given the inequality $a^2+b^2+c^2-ab-bc-ca \leq 0$. Multiplying by $2$ and rearranging,we get $(a-b)^2 + (b-c)^2 + (c-a)^2 \leq 0$. Since the sum of squares of real numbers is non-negative,this is only possible if $a-b=0$,$b-c=0$,and $c-a=0$,which implies $a=b=c$. Substituting $a=b=c$ into the determinant: $\left|\begin{array}{ccc} (a-a+1)^5 & a^7-a^7 & a^9-a^9 \ a^{11}-a^{11} & (a-a+2)^3 & a^{13}-a^{13} \ a^{15}-a^{15} & a^{17}-a^{17} & (a-a+3)^1 \end{array}ight|$ $= \left|\begin{array}{ccc} 1^5 & 0 & 0 \ 0 & 2^3 & 0 \ 0 & 0 & 3^1 \end{array}ight|$ $= \left|\begin{array}{ccc} 1 & 0 & 0 \ 0 & 8 & 0 \ 0 & 0 & 3 \end{array}ight| = 1 	imes 8 	imes 3 = 24$.

If $a, b$ and $c$ are real numbers such that $a^2+b^2+c^2-ab-bc-ac \leq 0$,then the value of the determinant $\left|\begin{array}{ccc} (a-b+1)^5 & b^7-c^7 & c^9-a^9 \\ a^{11}-b^{11} & (b-c+2)^3 & c^{13}-a^{13} \\ a^{15}-b^{15} & b^{17}-c^{17} & (c-a+3)^1 \end{array}\right|$ is:

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