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

Question

Aicharya · Accepted Answer

The given equations are $ \begin{aligned} & x-y+z=1, \ & x+y-z=-1 	ext { and } x-3 y+3 z=2 \end{aligned} $ The system of equations can be put in matrix form as $A X=B$ which is inconsistent as $ho(A: B) 
eq ho(A)$. $\Rightarrow$ The three planes do not have a common point. $\Rightarrow$ Statement 2 is true. Since, planes $P_1, P_2$ and $P_3$ are pairwise intersection, their lines of intersection are parallel. Statement 1 is false

Consider three planes $$ P_1: x-y+z=1, P_2: x+y-z=1 $$ and $P_3: x-3 y+3 z=2$ Let $L_1, L_2$ and $L_3$ be…

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