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(B) Let the lines be $L_1: 2x + 3y - 1 = 0$,$L_2: x + 2y + 1 = 0$,and $L_3: ax + by - 1 = 0$. The orthocenter is at $(0, 0)$.Since the orthocenter is

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$\frac{1}{60}$

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(B) Let the lines be $L_1: 2x + 3y - 1 = 0$,$L_2: x + 2y + 1 = 0$,and $L_3: ax + by - 1 = 0$. The orthocenter is at $(0, 0)$.Since the orthocenter is the intersection of altitudes,the altitude from vertex $B$ (intersection of $L_1$ and $L_2$) passes through $(0, 0)$ and is perpendicular to $L_3$. The family of lines through $B$ is $(2x + 3y - 1) + \lambda(x + 2y + 1) = 0$. Since it passes through $(0, 0)$,we have $-1 + \lambda = 0 \Rightarrow \lambda = 1$. The altitude is $3x + 5y = 0$,with slope $m = -\frac{3}{5}$. Since this is perpendicular to $L_3$ (slope $-\frac{a}{b}$),we have $(-\frac{a}{b}) 	imes (-\frac{3}{5}) = -1 \Rightarrow 3a = -5b$.Similarly,the altitude from vertex $A$ (intersection of $L_1$ and $L_3$) passes through $(0, 0)$ and is perpendicular to $L_2$ (slope $-\frac{1}{2}$). The family of lines through $A$ is $(2x + 3y - 1) + \mu(ax + by - 1) = 0$. Since it passes through $(0, 0)$,we have $-1 - \mu = 0 \Rightarrow \mu = -1$. The altitude is $(2-a)x + (3-b)y = 0$,with slope $-\frac{2-a}{3-b}$. Since this is perpendicular to $L_2$,we have $(-\frac{2-a}{3-b}) 	imes (-\frac{1}{2}) = -1$ $\Rightarrow 2-a = -2(3-b)$ $\Rightarrow a + 2b = 8$.Solving $3a = -5b$ and $a + 2b = 8$,we get $a = -40$ and $b = 24$. Wait,checking the calculation: $a = -40, b = 24 \Rightarrow 3(-40) = -120, -5(24) = -120$ (Correct). $a + 2b = -40 + 48 = 8$ (Correct). Thus,$\frac{1}{a} + \frac{1}{b} = -\frac{1}{40} + \frac{1}{24} = \frac{-3 + 5}{120} = \frac{2}{120} = \frac{1}{60}$.

If the orthocenter of the triangle formed by the lines $2x + 3y - 1 = 0$,$x + 2y + 1 = 0$,and $ax + by - 1 = 0$ lies at the origin,then $\frac{1}{a} + \frac{1}{b} =$

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