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If the points $(1,2,3)$ and $(2,-1,0)$ lie on the opposite sides of the plane $2 x+3 y-2 z=k$, then
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Verified Answer
The correct answer is:
$1 < k < 2$
The points $(1,2,3)$ and $(2,-1,0)$ lie on the opposite sides of the plane $2 x+3 y-2 z-k=0$
So, $(2+6-6-\mathrm{k})(4-3-\mathrm{k}) < 0$
$\begin{array}{l}
\Rightarrow(\mathrm{k}-1)(\mathrm{k}-2) < 0 \\
\therefore 1 < \mathrm{k} < 2
\end{array}$
So, $(2+6-6-\mathrm{k})(4-3-\mathrm{k}) < 0$
$\begin{array}{l}
\Rightarrow(\mathrm{k}-1)(\mathrm{k}-2) < 0 \\
\therefore 1 < \mathrm{k} < 2
\end{array}$
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