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JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant

જો સુરેખ સમીકરણો \(x - 2y + kz = 1\) ; \(2x + y + z = 2\) ;  \(3x - y - kz = 3\) નો ઉકેલ  \((x, y, z) \ne 0\),  હોય તો  \((x, y)\) એ  . .  . .  રેખા પર આવેલ છે .

  1. A \(3x -4y -1 = 0\)
  2. B \(4x -3y -4 = 0\)
  3. C \(4x -3y -1 = 0\)
  4. D \(3x -4y -4 = 0\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(4x -3y -4 = 0\)

Step-by-step Solution

Detailed explanation

For infinitly many solution \(\left| {\begin{array}{*{20}{c}} 1&{ - 2}&k\\ 2&1&1\\ 3&{ - 1}&{ - k} \end{array}} \right| - 0\) \( \Rightarrow k = \frac{{ - 1}}{2}\) Also consider \(x - 2y + k = 1\) and \(2x + y + z = 2\) \( \Rightarrow 2x - 4y - z - 2\) \(2x + y + z = 2\)…
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