Question

$A(2,6,2), B(-4,0, \lambda), C(2,3,-1)$ and $D(4,5,0),|\lambda| \leq 5$ are the vertices of a quadrilateral $A B C D$. If its area is $18$ square units, then $5-6 \lambda$ is equal to________

Answer 1

$ A (2,6,2) \quad B (-4,0, \lambda), C (2,3,-1) D (4,5,0) $
$ \text { Area }=\frac{1}{2}|\overrightarrow{B D} \times \overrightarrow{A C}|=18$
$\overrightarrow{A C} \times \overrightarrow{B D}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 0 & -3 & -3 \\ 8 & 5 & -\lambda\end{vmatrix}$
$ =(3 \lambda+15) \hat{i}-\hat{j}(-24)+\hat{k}(-24) $
$ \overrightarrow{A C} \times \overrightarrow{B D}=(3 \lambda+15) \hat{i}+24 \hat{j}-24 \hat{k} $
$ =\sqrt{(3 \lambda+15)^2+(24)^2+(24)^2}=36 $
$ =\lambda^2+10 \lambda+9=0 $
$ =\lambda=-1,-9 $
$ |\lambda| \leq 5 \Rightarrow \lambda=-1 $
$ 5-6 \lambda=5-6(-1)=11$