$$A(2, 6, 2), B(-4, 0, \lambda), C(2, 3, -1)$$ 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 _____.

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

Calculating the diagonal vectors $$\vec{AC}$$ and $$\vec{BD}$$:

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

$$\vec{BD} = (4 - (-4))\hat{i} + (5-0)\hat{j} + (0-\lambda)\hat{k} = 8\hat{i} + 5\hat{j} - \lambda\hat{k}$$

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

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

$$\text{Area} = \frac{1}{2} \vert{}\vec{AC} \times \vec{BD}\vert{} = 18 \implies \vert{}\vec{AC} \times \vec{BD}\vert{} = 36$$

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

$$9(\lambda + 5)^2 + 576 + 576 = 1296$$

$$9(\lambda + 5)^2 + 1152 = 1296 \implies 9(\lambda + 5)^2 = 144$$

$$(\lambda + 5)^2 = 16 \implies \lambda + 5 = \pm 4$$

$$\lambda_1 = -1, \quad \lambda_2 = -9$$

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