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

The area of a quadrilateral $$ABCD$$ in 3D space, defined by its vertices, can be found using the magnitude of the cross product of its diagonal vectors $$\vec{AC}$$ and $$\vec{BD}$$:

$$\text{Area} = \frac{1}{2} |\vec{AC} \times \vec{BD}|$$

$$\vec{AC} = \vec{C} - \vec{A} = (2 - 2)\hat{i} + (3 - 6)\hat{j} + (-1 - 2)\hat{k}$$

$$\vec{AC} = 0\hat{i} - 3\hat{j} - 3\hat{k}$$

$$\vec{BD} = \vec{D} - \vec{B} = (4 - (-4))\hat{i} + (5 - 0)\hat{j} + (0 - \lambda)\hat{k}$$

$$\vec{BD} = 8\hat{i} + 5\hat{j} - \lambda\hat{k}$$

$$\vec{AC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -3 & -3 \\ 8 & 5 & -\lambda \end{vmatrix}$$

$$\vec{AC} \times \vec{BD} = \hat{i}((-3)(-\lambda) - (-3)(5)) - \hat{j}(0(-\lambda) - (-3)(8)) + \hat{k}(0(5) - (-3)(8))$$

$$\vec{AC} \times \vec{BD} = (3\lambda + 15)\hat{i} - 24\hat{j} + 24\hat{k}$$

$$\frac{1}{2} |\vec{AC} \times \vec{BD}| = 18$$

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

$$\sqrt{(3\lambda + 15)^2 + (-24)^2 + (24)^2} = 36$$

$$(3\lambda + 15)^2 + 576 + 576 = 1296$$

$$(3\lambda + 15)^2 = 144$$

$$3\lambda + 15 = \pm 12$$

$$3\lambda = -15 \pm 12$$

$$\implies \lambda = -1, -9$$ 

$$\lambda = -1$$ as per the given condition

$$5 - 6(-1) = 5 + 6 = 11$$