Problem Statement
If four points with position vectors $\vec{a} = 6\hat{i} + 3\hat{j} - 4\hat{k}$, $\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k}$, $\vec{c} = 3\hat{i} + \lambda \hat{j} + 4\hat{k}$, and $\vec{d} = -4\hat{i} + 4\hat{j} - 4\hat{k}$ are coplanar, find the value of $\lambda$.
Verified Solution & Marking Scheme
Form three edge vectors emanating from point A
$$\vec{AB} = \vec{b} - \vec{a} = (1 - 6)\hat{i} + (-2 - 3)\hat{j} + (3 - (-4))\hat{k} = -5\hat{i} - 5\hat{j} + 7\hat{k}$$
$$\vec{AC} = \vec{c} - \vec{a} = (3 - 6)\hat{i} + (\lambda - 3)\hat{j} + (4 - (-4))\hat{k} = -3\hat{i} + (\lambda - 3)\hat{j} + 8\hat{k}$$
$$\vec{AD} = \vec{d} - \vec{a} = (-4 - 6)\hat{i} + (4 - 3)\hat{j} + (-4 - (-4))\hat{k} = -10\hat{i} + \hat{j} + 0\hat{k}$$
Apply the Coplanarity Condition [AB AC AD] = 0
$$\begin{vmatrix} -5 & -5 & 7 \\ -3 & \lambda - 3 & 8 \\ -10 & 1 & 0 \end{vmatrix} = 0$$
Expand along the third row (which has a zero):
$$-10\Big( (-5)(8) - 7(\lambda - 3) \Big) - 1\Big( (-5)(8) - 7(-3) \Big) + 0 = 0$$
$$-10(-40 - 7\lambda + 21) - 1(-40 + 21) = 0$$
$$-10(-19 - 7\lambda) - (-19) = 0$$
$$190 + 70\lambda + 19 = 0 \implies 70\lambda = -209 \implies \lambda = -\frac{209}{70}$$