Question:

<p><strong>Step 1:</strong> Rewrite circle in standard form: $(x-3)^2 + (y-2)^2 = 25$. Center C(3,2), radius = 5.</p><p><strong>Step 2:</strong> Let chord pass through origin with equation $y = mx$. Find intersection points P and Q with circle by substituting into $x^2 + y^2 - 6x - 4y - 12 = 0$:</p><p>$x^2 + m^2x^2 - 6x - 4mx - 12 = 0$</p><p>$x^2(1+m^2) - 2x(3+2m) - 12 = 0$</p><p><strong>Step 3:</strong> If O divides PQ in ratio 3:2, then $\frac{OP}{OQ} = \frac{3}{2}$. If P corresponds to parameter $x_1$ and Q to $x_2$, then $\frac{|x_1|}{|x_2|} = \frac{3}{2}$ (considering signed lengths from O).</p><p><strong>Step 4:</strong> From Vieta's formulas: $x_1 + x_2 = \frac{2(3+2m)}{1+m^2}$ and $x_1 x_2 = \frac{-12}{1+m^2}$</p><p>For ratio 3:2 with opposite signs: $x_1 = -\frac{3k}{2}$ and $x_2 = k$ for some $k > 0$</p><p><strong>Step 5:</strong> Then $x_1 x_2 = -\frac{3k^2}{2} = \frac{-12}{1+m^2}$ gives $k^2 = \frac{8}{1+m^2}$</p><p>And $x_1 + x_2 = -\frac{k}{2} = \frac{2(3+2m)}{1+m^2}$</p><p>Solving these simultaneously: $m = 2$ or $m = -\frac{1}{2}$</p><p><strong>Step 6:</strong> The chords are $y = 2x$ and $2y + x = 0$ (or $x + 2y = 0$).</p><p>∴ Answer: A</p>