Three lines are given by $\vec{r} = \lambda\hat{i}, \lambda \in R$ ; $\vec{r} = \mu(\hat{i} + \hat{j}), \mu \in R$ and $\vec{r} = \nu(\hat{i} + \hat{j} + \hat{k}), \nu \in R$. 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 _____.

$L_1: \vec r = \lambda\hat i$

$L_2: \vec r = \mu(\hat i+\hat j)$

$L_3: \vec r = \nu(\hat i+\hat j+\hat k)$

$\pi: x+y+z=1$

$A = L_1\cap\pi,\ B = L_2\cap\pi,\ C = L_3\cap\pi$

$\text{To find: } (6\Delta)^2$

$\pi: x+y+z=1 \text{ cut by } L_1, L_2, L_3$ $1.\ \text{each line } \cap \pi \Rightarrow A, B, C$ $2.\ \text{vectors } \vec{AB}, \vec{AC}$ $3.\ \Delta = \tfrac12|\vec{AB}\times\vec{AC}|$ $L_1:\ (\lambda,0,0);\quad \lambda = 1$ $A=(1,0,0)$ $L_2:\ (\mu,\mu,0);\quad 2\mu = 1 \Rightarrow \mu=\tfrac12$ $B=\left(\tfrac12,\tfrac12,0\right)$ $L_3:\ (\nu,\nu,\nu);\quad 3\nu=1 \Rightarrow \nu=\tfrac13$ $C=\left(\tfrac13,\tfrac13,\tfrac13\right)$ $A, B, C \in \pi$ $\Delta = \tfrac12\,|\vec{AB}\times\vec{AC}|$ $\vec{AB} = B-A = \left(-\tfrac12,\ \tfrac12,\ 0\right)$ $\vec{AC} = C-A = \left(-\tfrac23,\ \tfrac13,\ \tfrac13\right)$ $\vec{AB}\times\vec{AC} = \begin{vmatrix} \hat i & \hat j & \hat k \\ -\tfrac12 & \tfrac12 & 0 \\ -\tfrac23 & \tfrac13 & \tfrac13 \end{vmatrix}$ $= \left(\tfrac16,\ \tfrac16,\ \tfrac16\right)$ $|\vec{AB}\times\vec{AC}| = \sqrt{3\cdot\tfrac{1}{36}} = \dfrac{\sqrt3}{6}$ $\Delta = \tfrac12\cdot\dfrac{\sqrt3}{6} = \dfrac{\sqrt3}{12}$ $6\Delta = 6\cdot\dfrac{\sqrt3}{12} = \dfrac{\sqrt3}{2}$ $(6\Delta)^2 = \left(\dfrac{\sqrt3}{2}\right)^2 = \dfrac34$ $\boxed{(6\Delta)^2 = \dfrac34}$