Consider three planes $P_1 : x - y + z = 1$; $P_2 : x + y - z = -1$ $P_3 : x - 3y + 3z = 2$ Let $L_1, L_2, L_3$ be the lines of intersection of the planes $P_2$ and $P_3$, $P_3$ and $P_1$, $P_1$ and $P_2$, respectively. Statement-I : At least two of the lines $L_1, L_2$ and $L_3$ are non-parallel Statement-II : The three planes do not have a common point.

Consider three planes P₁ : x - y + z = 1; P₂ : x + y - z = -1 P₃ : x - 3y + 3z = 2 Let L₁, L₂, L₃ be the lines of intersection of the planes P₂ and P₃, P₃ and P₁, P₁ and P₂, respectively. Statement-I : At least two of the lines L₁, L₂ and L₃ are non-parallel Statement-II : The three planes do not have a common point.

$P_1: x - y + z = 1, \qquad P_2: x + y - z = -1$

$P_3: x - 3y + 3z = 2$

$L_1 = P_2 \cap P_3, \quad L_2 = P_3 \cap P_1, \quad L_3 = P_1 \cap P_2$

$\text{Check: S-I (non-parallel?) and S-II (common point?)}$

$1.\ \text{normals } \vec{n}_1, \vec{n}_2, \vec{n}_3$

$2.\ \vec{d}_{L_i} = \vec{n}_j \times \vec{n}_k \Rightarrow \text{S-I}$

$3.\ \text{consistency of } P_1, P_2, P_3 \Rightarrow \text{S-II}$

$P_1: \vec{n}_1 = (1, -1, 1), \qquad P_2: \vec{n}_2 = (1, 1, -1)$

$P_3: \vec{n}_3 = (1, -3, 3)$

$\vec{d}_{L_1} = \vec{n}_2 \times \vec{n}_3 = (0, -4, -4)$

$\vec{d}_{L_2} = \vec{n}_3 \times \vec{n}_1 = (0, 2, 2)$

$\vec{d}_{L_3} = \vec{n}_1 \times \vec{n}_2 = (0, 2, 2)$

$(0, -4, -4) \parallel (0, 2, 2) \parallel (0, 2, 2) \parallel (0, 1, 1)$

$\Rightarrow L_1, L_2, L_3 \text{ all parallel} \Rightarrow \text{S-I is FALSE}$

$\vec{n}_3 = 2\vec{n}_1 - \vec{n}_2$

$2P_1 - P_2: \ 2(x-y+z) - (x+y-z) = 2(1) - (-1)$

$\Rightarrow x - 3y + 3z = 3 \ \neq \ 2$

$\text{common point} \Rightarrow x - 3y + 3z = 3$

$\text{but } P_3: x - 3y + 3z = 2 \Rightarrow \text{contradiction}$

$\Rightarrow \text{no common point} \Rightarrow \text{S-II is TRUE}$

$P_1 + P_2: \ 2x = 0 \Rightarrow x = 0$

$x = 0 \Rightarrow P_1: -y + z = 1, \quad P_3: -y + z = \tfrac{2}{3}$

$1 \neq \tfrac{2}{3} \Rightarrow \text{inconsistent}$

$\text{S-I is FALSE, S-II is TRUE}$

$\Rightarrow \text{option (4)}$