\begin{aligned} &l_{1}: a x+2 b y+3 c=0 ; \\ &l_{2}: b x+2 c y+3 a=0 ; \\ &l_{3}: c x+2 a y+3 b=0 . \end{aligned}

$$\left\{\begin{array}{l} a x+2 b y=-3 c \\ b x+2 c y=-3 a, \\ c x+2 a y=-3 b \end{array}\right.$$

\begin{aligned} |\bar{A}| &=\left|\begin{array}{ccc} a & 2 b & -3 c \\ b & 2 c & -3 a \\ c & 2 a & -3 b \end{array}\right|=\left|\begin{array}{ccc} a+b+c & 2(b+c+a) & -3(c+a+b) \\ b & 2 c & -3 a \\ c & 2 a & -3 b \end{array}\right| \\ &=(a+b+c)\left|\begin{array}{ccc} 1 & 2 & -3 \\ b & 2 c & -3 a \\ c & 2 a & -3 b \end{array}\right|=-6(a+b+c)\left|\begin{array}{ccc} 1 & 1 & 1 \\ b & c & a \\ c & a & b \end{array}\right| \end{aligned}

\begin{aligned} &=-6(a+b+c)\left|\begin{array}{ccc} 1 & 0 & 0 \\ b & c-b & a-b \\ c & a-c & b-c \end{array}\right|=-6(a+b+c)\left|\begin{array}{cc} c-b & a-b \\ a-c & b-c \end{array}\right| \\ &=-6(a+b+c)[(c-b)(b-c)-(a-b)(a-c)] \\ &=-6(a+b+c)\left(b c-c^{2}-b^{2}+b c-a^{2}+a c+a b-b c\right) \\ &=6(a+b+c)\left(a^{2}+b^{2}+c^{2}-a c-a b-b c\right) \\ &=3(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right], \end{aligned}

“充分性”. 由 $a+b+c=0$, 则从必要性的证明可知, $|\bar{A}|=0$, 故秩 $(\bar{A}) < 3$.

$$\left|\begin{array}{ll} a & 2 b \\ b & 2 c \end{array}\right|=2\left(a c-b^{2}\right)=-2\left[a(a+b)+b^{2}\right]=-2\left[\left(a+\frac{1}{2} b\right)^{2}+\frac{3}{4} b^{2}\right] \neq 0,$$

$$\text { 设三直线交于一点 }\left(x_{0}, y_{0}\right) \text {, 则 }\left[\begin{array}{c} x_{0} \\ y_{0} \\ 1 \end{array}\right] \text { 为 } B X=0 \text { 的非零解, 其中 } B=\left[\begin{array}{ccc} a & 2 b & 3 c \\ b & 2 c & 3 a \\ c & 2 a & 3 b \end{array}\right] \text {. }$$

\begin{aligned} |B| &=\left|\begin{array}{lll} a & 2 b & 3 c \\ b & 2 c & 3 a \\ c & 2 a & 3 b \end{array}\right|=-\left|\begin{array}{lll} a & 2 b & -3 c \\ b & 2 c & -3 a \\ c & 2 a & -3 b \end{array}\right|=-|\bar{A}| \\ &=-3(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right],(\text { 解法同方法 1) } \end{aligned}

“充分性”: 考虑线性方程组
$$\left\{\begin{array}{l} a x+2 b y=-3 c \\ b x+2 c y=-3 a \\ c x+2 a y=-3 b \end{array}\right.$$

$$\left\{\begin{array}{l} a x+2 b y=-3 c, \\ b x+2 c y=-3 a . \end{array}\right.$$

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