# The following system of linear equations 2x + 3y + 2z = 9 3x + 2y + 2z = 9 x - y + 4z = 8   Option: 1 has a solution $(\alpha , \beta ,\gamma )$ satisfying $\alpha + \beta ^2 + \gamma ^3 = 12$ Option: 2 has infinitely many solutions   Option: 3 does not have any solution   Option: 4 has a unique solution

$\\2 x+3 y+2 z=9\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(1) \\ 3 x+2 y+2 z=9 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(2)\\ x-y+4 z=8\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(3)$

(1) – (2)

⇒ –x + y = 0

⇒ x – y = 0

from (3)

4z = 8

⇒ z = 2

from (1)

2x + 3y = 5

⇒ x = y = 1

therefore, system has unique solution

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