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(D) Let the lines be $y - mx = 0$ and $y - m_1x = 0$ for the first equation,and $y - mx = 0$ and $y - m_2x = 0$ for the second equation,where $y - mx

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$6$

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(D) Let the lines be $y - mx = 0$ and $y - m_1x = 0$ for the first equation,and $y - mx = 0$ and $y - m_2x = 0$ for the second equation,where $y - mx = 0$ is the common line. For $2x^2 + \alpha xy + 3y^2 = 0$,we have $3y^2 + \alpha xy + 2x^2 = 0$. Dividing by $x^2$,we get $3(y/x)^2 + \alpha(y/x) + 2 = 0$. Let $m$ and $m_1$ be the roots. Then $m + m_1 = -\alpha/3$ and $m \cdot m_1 = 2/3$. For $2x^2 + \beta xy - 3y^2 = 0$,we have $-3y^2 + \beta xy + 2x^2 = 0$,or $3y^2 - \beta xy - 2x^2 = 0$. Dividing by $x^2$,we get $3(y/x)^2 - \beta(y/x) - 2 = 0$. Let $m$ and $m_2$ be the roots. Then $m + m_2 = \beta/3$ and $m \cdot m_2 = -2/3$. Given that the other two lines $y - m_1x = 0$ and $y - m_2x = 0$ are perpendicular,$m_1 \cdot m_2 = -1$. From $m \cdot m_1 = 2/3$,$m_1 = 2/(3m)$. From $m \cdot m_2 = -2/3$,$m_2 = -2/(3m)$. Substituting into $m_1 \cdot m_2 = -1$: $(2/(3m)) \cdot (-2/(3m)) = -1$ $\Rightarrow -4/(9m^2) = -1$ $\Rightarrow m^2 = 4/9$ $\Rightarrow m = \pm 2/3$. If $m = 2/3$,$m_1 = 1$ and $m_2 = -1$. Then $\alpha = -3(m + m_1) = -3(2/3 + 1) = -5$ and $\beta = 3(m + m_2) = 3(2/3 - 1) = -1$. $|\alpha + \beta| = |-5 - 1| = 6$. If $m = -2/3$,$m_1 = -1$ and $m_2 = 1$. Then $\alpha = -3(-2/3 - 1) = 5$ and $\beta = 3(-2/3 + 1) = 1$. $|\alpha + \beta| = |5 + 1| = 6$.

The numbers $\alpha$ and $\beta$ are such that one of the lines of $2x^2 + \alpha xy + 3y^2 = 0$ coincides with one of the lines of $2x^2 + \beta xy - 3y^2 = 0$. If the two lines other than that common line are perpendicular,then $|\alpha + \beta|$ is equal to

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