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The system of equations 2x+y+kz=1,12y+3z=2,  x+y2z=3 is consistent if

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a
a+b+c=0
b
a+b+c≤0
c
k≠−1
d
a+b+c≥0

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detailed solution

Correct option is C

-21k1-2311-2≠0⇒-2[1]-[-5]+k[3]≠0⇒-2+5+3k≠0⇒3k+3≠0⇒k≠-1


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