A (2, 6, 2), B (- 4, 0, λ), C(2, 3, - 1) and D(4, 5, 0), |λ| ≤ 5...

Given:

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

Area of quadrilateral

$= \dfrac{1}{2}\left|\overset{\to}{AC} \times \overset{\to}{BD}\right| = 18$

$⇒ \left|\overset{\to}{AC} \times \overset{\to}{BD}\right|36$

Now,

\overset{\to}{AC} \times \overset{\to}{BD} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 0 & - 3 & - 3 \\ 8 & 5 & - \lambda\end{cases}\right|

⇒ \overset{\to}{AC} \times \overset{\to}{BD} = \left(3\lambda + 15\right)\overset{^}{i} - \left(0 + 24\right)\overset{^}{j} + \left(0 + 24\right)\overset{^}{k}

⇒ \overset{\to}{AC} \times \overset{\to}{BD} = \left(3\lambda + 15\right)\overset{^}{i} - 24\overset{^}{j} + 24\overset{^}{k}

So,

\left|\overset{\to}{AC} \times \overset{\to}{BD}\right| = 36$</p><p>$⇒ \sqrt{{\left(3\lambda + 15\right)}^{2} + {\left(24\right)}^{2} + {\left(24\right)}^{2}} = 36$</p><p>$⇒ 9{\lambda}^{2} + 225 + 90\lambda = 1296 - 1152$</p><p>$⇒ 9{\lambda}^{2} + 90\lambda - 81 = 0$</p><p>$= {\lambda}^{2} + 10\lambda + 9 = 0$</p><p>$⇒ \lambda = - 1, - 9$</p><p>Since,</p><p>\left|\lambda\right| \le 5 ⇒ \lambda = - 1$

Therefore,

$5 - 6\lambda = 5 - 6\left(- 1\right) = 11$