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(C) Given the equation: $l(3a + 2b + c) + m(2a + 2b + 3c) + n(a + 2b + 5c) = 0$ Rearranging the terms based on the vectors $a, b, c$: $a(3l + 2m + n)

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$l = n, m + 2n = 0$

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(C) Given the equation: $l(3a + 2b + c) + m(2a + 2b + 3c) + n(a + 2b + 5c) = 0$ Rearranging the terms based on the vectors $a, b, c$: $a(3l + 2m + n) + b(2l + 2m + 2n) + c(l + 3m + 5n) = 0$ Since $a, b, c$ are linearly independent, the coefficients must be zero: $3l + 2m + n = 0$  $(i)$ $2l + 2m + 2n = 0 \Rightarrow l + m + n = 0$  $(ii)$ $l + 3m + 5n = 0$  $(iii)$ Subtracting $(ii)$ from $(i)$: $(3l + 2m + n) - (l + m + n) = 0 \Rightarrow 2l + m = 0 \Rightarrow m = -2l$ Subtracting $(ii)$ from $(iii)$: $(l + 3m + 5n) - (l + m + n) = 0 \Rightarrow 2m + 4n = 0 \Rightarrow m = -2n$ Equating the two expressions for $m$: $-2l = -2n \Rightarrow l = n$ Substituting $l = n$ into $m = -2n$: $m = -2n \Rightarrow m + 2n = 0$ Thus, the condition is $l = n$ and $m + 2n = 0$.

If $a, b, c$ are three linearly independent vectors and there exists a non-zero scalar triad $(l, m, n)$ such that $l(3a + 2b + c) + m(2a + 2b + 3c) + n(a + 2b + 5c) = 0$, then:

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