Q56 / 120 · VITEEE 2014 · Mathematics

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

  • A
    $k < 1$
  • B
    $k>2$
  • C
    $k < 1$ or $k>2$
  • Correct answer
    D
    $1 < k < 2$

Solution

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, $(2+6-6-\mathrm{k})(4-3-\mathrm{k}) < 0$ $\begin{array}{l} \Rightarrow(\mathrm{k}-1)(\mathrm{k}-2) < 0 \\ \therefore 1 < \mathrm{k} < 2 \end{array}$

Hence, the correct answer is option: 'D'.

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