1) 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 D) k < 1 or k > 2 E) 1 < k < 2 Answer: Option DExplanation:The points (1,2,3) and (2,-1,0) lie on the opposite sides of the plane 2x+3y-2z-k=0 so, (2+6-6-k)(4-3-k)<0 $\Rightarrow$ $(k-1)(k-2)<0$ $\therefore$ $1<k<2$
