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JEE Advanced · Mathematics · 19. Determinants

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.

  1. A
    Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.
  2. B
    Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
  3. C
    Statement 1 is true, Statement 2 is false.
  4. D
    Statement 1 is false, Statement 2 is true
Verified Solution

Answer & Solution

Correct Answer

(D)
Statement 1 is false, Statement 2 is true

Step-by-step Solution

Detailed explanation

The given equations are
\[
\begin{aligned}
& x-y+z=1, \\
& x+y-z=-1 \text { and } x-3 y+3 z=2
\end{aligned}
\]
The system of equations can be put in matrix form as \(A X=B\)


which is inconsistent as \(\rho(A: B) \neq \rho(A)\).
\(\Rightarrow\) 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.
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