A) \[\frac{1}{7}\]
B) 16
C) 7
D) 12
Correct Answer: C
Solution :
Point \[(2,-3)\] lies on the line \[k{{x}^{2}}-3{{y}^{2}}+2x+y-2=0\] Þ \[{{(2)}^{2}}k-3{{(-3)}^{2}}+2.2-3-2=0\] \[4k-28=0\Rightarrow k=7\].
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