The system of equations $x + 2 y + 3 z = 4, 2 x + 3 y + 4 z = 5, 3 x + 4 y + 5 z = 6$ has

infinitely many solution

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no solution

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unique solution

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none of these

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Solution

The correct option is A infinitely many solution
Consider
$3 x + 4 y + 5 z = 6$ ...(i)
$2 x + 3 y + 4 z = 5$ ...(ii)
$x + 2 y + 3 z = 4$ ...(iii)
Therefore, $∣ ∣ ∣ ∣ \begin{matrix} 3 & 4 & 5 2 & 3 & 4 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣$ $= 3 (9 - 8) - 4 (6 - 4) + 5 (4 - 3)$
$= 3 + 5 - (4 \times 2)$
$= 8 - 8$
$= 0$
Hence the above set of equations have infinitely number of planes.

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x + 2 y + 3 z = 4, 2 x + 3 y + 4 z = 5, 3 x + 4 y + 5 z = 6

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