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If $a, b$ and $c$ are real numbers such that $a^2+b^2+c^2-a b-b c-a c \leq 0$, then
$$
\left|\begin{array}{ccc}
(a-b+1)^5 & b^7-c^7 & c^9-a^9 \\
a^{11}-b^{11} & (b-c+2)^3 & c^{13}-a^{13} \\
a^{15}-b^{15} & b^{17}-c^{17} & (c-a+3)^1
\end{array}\right|=
$$
Options:
$$
\left|\begin{array}{ccc}
(a-b+1)^5 & b^7-c^7 & c^9-a^9 \\
a^{11}-b^{11} & (b-c+2)^3 & c^{13}-a^{13} \\
a^{15}-b^{15} & b^{17}-c^{17} & (c-a+3)^1
\end{array}\right|=
$$
Solution:
1827 Upvotes
Verified Answer
The correct answer is:
24
Given, $a^2+b^2+c^2-a b-b c-c a \leq 0$
$$
\therefore \frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \leq 0
$$
It is possible only $a=b=c$
$$
\begin{aligned}
& \text { Now, }\left|\begin{array}{ccc}
(a-b+1)^5 & b^7-c^7 & c^9-a^9 \\
a^{11}-b^{11} & (b-c+2)^3 & c^{13}-a^{13} \\
a^{15}-b^{15} & b^{17}-c^{17} & (c-a+3) 1
\end{array}\right| \\
& \Rightarrow\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 8 & 0 \\
0 & 0 & 3
\end{array}\right|=24
\end{aligned}
$$
$$
\therefore \frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \leq 0
$$
It is possible only $a=b=c$
$$
\begin{aligned}
& \text { Now, }\left|\begin{array}{ccc}
(a-b+1)^5 & b^7-c^7 & c^9-a^9 \\
a^{11}-b^{11} & (b-c+2)^3 & c^{13}-a^{13} \\
a^{15}-b^{15} & b^{17}-c^{17} & (c-a+3) 1
\end{array}\right| \\
& \Rightarrow\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 8 & 0 \\
0 & 0 & 3
\end{array}\right|=24
\end{aligned}
$$
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