Question

If the point $A(2,-9, \lambda)$ and $B(\lambda,-1,-3)$ lie on the opposite sides of a plane which contain the lines $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and $\frac{x}{1}=\frac{7-y}{3}=\frac{z+7}{2}$, then find the largest integral value of $\lambda$.

Answer 1

$\because\begin{vmatrix}-1-0 & 3-7 & -2+7 \\ -3 & 2 & 1 \\ 1 & 3 & 2\end{vmatrix}=0$
$\Rightarrow$ lines are intersecting and for intersection points
$-3 r_1-1=r_2 $ .....(1)
$2 r_1+3=7-3 r_2$ .....(2)
$-2+r_1=2 r_2-7 $ .....(3)
$\Rightarrow r_1=-1, r_2=-2$
$\therefore$ intersection point is $(2,1,-3)$
$\therefore$ Equation of plane is $\begin{vmatrix}x-2 & y-1 & z+3 \\ -3 & 2 & 1 \\ 1 & 3 & 2\end{vmatrix}=0$
$\Rightarrow x+y+z=0 $
$\because A (2,-9, \lambda) \text { and } B (\lambda,-1,-3) \text { lie on the opposite side of the of the plane } $
$\Rightarrow (2-9+\lambda)(\lambda-1-3)<0 $
$(\lambda-4)(\lambda-7)<0 $
$\Rightarrow \text { Largest integral value is } \lambda=6 $