Question

If the points $(1,2,3)$ and $(2,-1,0)$ lie on the opposite sides of the plane $2x+3y-2z=k,$ then

• A. $k<1$
• B. $k>2$
• C. $k<1ork>2$
• D. $1<k<2$

Answer 1

The correct option is D

$1<k<2$

Find the range of $\mathit{k}$:

Given, that the points $(1,2,3)$ and $(2,-1,0)$lie on the opposite sides of the plane $2x+3y-2z-k=0$

$\Rightarrow$ $(2+6-6-k)(4-3-0-k)<0$

$\Rightarrow$$\left(2-k\right)\left(1-k\right)<0$ $....\left(i\right)$

For $k<1$

$(2-k)>0,(1-k)>0\Rightarrow \left(2-k\right)\left(1-k\right)>0$

For $k>2$

$(2-k)<0,(1-k)<0\Rightarrow \left(2-k\right)\left(1-k\right)>0$

For $1<k<2$

$(2-k)>0,(1-k)<0\Rightarrow \left(2-k\right)\left(1-k\right)<0$

Therefore, $1<k<2$ is the range of values of $k$ for which inequality $\left(i\right)$ satisfied.

Hence, option D is correct option.