A(2,6,2),B(-4,0,λ ),C(2,3,-1) and D(4,5,0),|λ |≤ 5 are the vertices of a quadrilateral ABCD. If its area is 18 square units, then 5-6λ is equal to .

Given:

$$A(2,6,2),B(-4,0,\lambda ),C(2,3,-1)D(4,5,0)$$A(2,6,2),B(−4,0,λ),C(2,3,−1)D(4,5,0)

Area of quadrilateral

$$=\frac{1}{2}|\vec{AC}\times \vec{BD}|=18$$=21​∣AC×BD∣=18

$$\Rightarrow |\vec{AC}\times \vec{BD}|36$$⇒∣AC×BD∣36

Now,

$$\vec{AC}\times \vec{BD}=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 0 & -3 & -3 \\ 8 & 5 & -\lambda \end{matrix}|$$AC×BD=∣i^08​j^​−35​k^−3−λ​∣

$$\Rightarrow \vec{AC}\times \vec{BD}=(3\lambda +15)\hat{i}-(0+24)\hat{j}+(0+24)\hat{k}$$⇒AC×BD=(3λ+15)i^−(0+24)j^​+(0+24)k^

$$\Rightarrow \vec{AC}\times \vec{BD}=(3\lambda +15)\hat{i}-24\hat{j}+24\hat{k}$$⇒AC×BD=(3λ+15)i^−24j^​+24k^

So,

$$|\vec{AC}\times \vec{BD}|=36$$∣AC×BD∣=36

$$\Rightarrow \sqrt{{(3\lambda +15)}^{2}+{(24)}^{2}+{(24)}^{2}}=36$$⇒(3λ+15)2+(24)2+(24)2​=36

$$\Rightarrow 9{\lambda }^{2}+225+90\lambda =1296-1152$$⇒9λ2+225+90λ=1296−1152

$$\Rightarrow 9{\lambda }^{2}+90\lambda -81=0$$⇒9λ2+90λ−81=0

$$={\lambda }^{2}+10\lambda +9=0$$=λ2+10λ+9=0

$$\Rightarrow \lambda =-1,-9$$⇒λ=−1,−9

Since,

$$|\lambda |\leq 5\Rightarrow \lambda =-1$$∣λ∣≤5⇒λ=−1

Therefore,

$$5-6\lambda =5-6(-1)=11$$5−6λ=5−6(−1)=11