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(A) The family of lines $(x - 2y - 3) + \lambda (x + 3y + 2) = 0$ passes through the intersection of $x - 2y - 3 = 0$ and $x + 3y + 2 = 0$.Solving the

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$x + 4y + 3 = 0$

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(A) The family of lines $(x - 2y - 3) + \lambda (x + 3y + 2) = 0$ passes through the intersection of $x - 2y - 3 = 0$ and $x + 3y + 2 = 0$.Solving these equations: $x - 2y = 3$ and $x + 3y = -2$.Subtracting the first from the second gives $5y = -5$,so $y = -1$.Substituting $y = -1$ into $x - 2y = 3$ gives $x - 2(-1) = 3$,so $x = 1$.The fixed point is $P(1, -1)$.The lines $L_1$ and $L_2$ pass through $P(1, -1)$.The bisector $B_1$ passes through $P(1, -1)$ and $A(2, 3)$.The slope of $B_1$ is $m_1 = \frac{3 - (-1)}{2 - 1} = \frac{4}{1} = 4$.The equation of $B_1$ is $y - 3 = 4(x - 2)$,which simplifies to $4x - y - 5 = 0$.The angle bisectors of two lines are always perpendicular to each other.Let the slope of the other bisector $B_2$ be $m_2$. Since $B_1 \perp B_2$,$m_1 	imes m_2 = -1$,so $4 	imes m_2 = -1$,which gives $m_2 = -\frac{1}{4}$.The equation of $B_2$ passing through $P(1, -1)$ is $y - (-1) = -\frac{1}{4}(x - 1)$.$4(y + 1) = -(x - 1)$ $\Rightarrow 4y + 4 = -x + 1$ $\Rightarrow x + 4y + 3 = 0$.

Let distinct lines $L_1$ and $L_2$ belong to the family of lines $(x - 2y - 3) + \lambda (x + 3y + 2) = 0$. If $B_1$ is the angle bisector of $L_1$ and $L_2$ which passes through the point $A(2, 3)$,then the equation of the other bisector of $L_1$ and $L_2$ is ($\lambda$ is a parameter).

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