For a, b, c in R,if 6 a^2-3 b^2-c^2+7 a b-a c | vedclass

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(C) Given equation is $6 a^2-3 b^2-c^2+7 a b-a c+4 b c=0$. Rearranging as a quadratic in $a$: $6 a^2 + a(7b - c) - (3b^2 - 4bc + c^2) = 0$. Factorizin

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concurrent at $(-2,-3)$ or $(3,-1)$

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(C) Given equation is $6 a^2-3 b^2-c^2+7 a b-a c+4 b c=0$. Rearranging as a quadratic in $a$: $6 a^2 + a(7b - c) - (3b^2 - 4bc + c^2) = 0$. Factorizing the quadratic expression: $6 a^2 + a(7b - c) - (3b - c)(b - c) = 0$. $6 a^2 + 9ab - 6ac - 2ab + 2ac - (3b - c)(b - c) = 0$. $(3a - b + c)(2a + 3b - c) = 0$. This implies either $3a - b + c = 0$ or $2a + 3b - c = 0$. Case $1$: $3a - b + c = 0$. Comparing with $ax + by + c = 0$,we get $x = 3, y = -1$. Case $2$: $2a + 3b - c = 0$. Comparing with $ax + by + c = 0$,we get $x = 2, y = -3$. Thus,the lines are concurrent at $(3, -1)$ or $(2, -3)$.

Column $I$	Column $II$
$(A)$ $L_1, L_2, L_3$ are concurrent,if	$(p)$ $k=-9$
$(B)$ One of $L_1, L_2, L_3$ is parallel to at least one of the other two,if	$(q)$ $k=-\frac{6}{5}$
$(C)$ $L_1, L_2, L_3$ form a triangle,if	$(r)$ $k=\frac{5}{6}$
$(D)$ $L_1, L_2, L_3$ do not form a triangle,if	$(s)$ $k=5$

For $a, b, c \in R$,if $6 a^2-3 b^2-c^2+7 a b-a c+4 b c=0$ and $|a|+|b| \neq 0$,then all the lines given by $a x+b y+c=0$ are

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