Three lines are given by $\vec{r} = \lambda \hat{i}, \lambda \in R$; $\vec{r} = \mu(\hat{i} + \hat{j}), \mu \in R$ and $\vec{r} = v(\hat{i} + \hat{j} + \hat{k}), v \in R$. Let the lines cut the plane $x + y + z = 1$ at the points A, B and C respectively. If the area of the triangle ABC is $\Delta$ then the value of $(6\Delta)^2$ equals ____.

$\text{Given: } L_1: \vec{r} = \lambda\hat{i}$ $L_2: \vec{r} = \mu(\hat{i}+\hat{j})$ $L_3: \vec{r} = v(\hat{i}+\hat{j}+\hat{k})$ $\text{Cuts } x+y+z=1 \text{ at } A,\ B,\ C \text{ respectively}$ $\text{To find: } (6\Delta)^2, \quad \Delta = [ABC]$ $\text{Plan: } 1.\ L_i \cap (x+y+z=1) \Rightarrow A,\ B,\ C$ $2.\ \overrightarrow{AB},\ \overrightarrow{AC}$ $3.\ \Delta = \tfrac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}| \Rightarrow (6\Delta)^2$ $L_1: (\lambda, 0, 0); \quad \lambda + 0 + 0 = 1 \Rightarrow \lambda = 1$ $A = (1,\ 0,\ 0)$ $L_2: (\mu, \mu, 0); \quad 2\mu = 1 \Rightarrow \mu = \tfrac{1}{2}$ $B = \left(\tfrac{1}{2},\ \tfrac{1}{2},\ 0\right)$ $L_3: (v, v, v); \quad 3v = 1 \Rightarrow v = \tfrac{1}{3}$ $C = \left(\tfrac{1}{3},\ \tfrac{1}{3},\ \tfrac{1}{3}\right)$ $\overrightarrow{AB} = B - A = \left(\tfrac{1}{2}-1,\ \tfrac{1}{2}-0,\ 0\right)$ $\overrightarrow{AB} = \left(-\tfrac{1}{2},\ \tfrac{1}{2},\ 0\right)$ $\overrightarrow{AC} = C - A = \left(\tfrac{1}{3}-1,\ \tfrac{1}{3}-0,\ \tfrac{1}{3}-0\right)$ $\overrightarrow{AC} = \left(-\tfrac{2}{3},\ \tfrac{1}{3},\ \tfrac{1}{3}\right)$ $\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -\tfrac{1}{2} & \tfrac{1}{2} & 0 \\ -\tfrac{2}{3} & \tfrac{1}{3} & \tfrac{1}{3} \end{vmatrix}$ $= \left(\tfrac{1}{6},\ \tfrac{1}{6},\ \tfrac{1}{6}\right)$ $|\overrightarrow{AB}\times\overrightarrow{AC}| = \sqrt{\tfrac{1}{36}+\tfrac{1}{36}+\tfrac{1}{36}} = \dfrac{\sqrt{3}}{6}$ $\Delta = \tfrac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}| = \dfrac{\sqrt{3}}{12}$ $6\Delta = 6\cdot\dfrac{\sqrt{3}}{12} = \dfrac{\sqrt{3}}{2}$ $(6\Delta)^2 = \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3}{4}$ $(6\Delta)^2 = \dfrac{3}{4}$