If the line $$y = mx + 1$$ meets the circle $$x^2 + y^2 + 3x = 0$$ in two points equidistant from and on opposite sides of $$x$$-axis, then

The two intersection points of the line $$y = mx + 1$$ with the circle $$x^2 + y^2 + 3x = 0$$ must have equal magnitudes of the $$y$$-coordinate but opposite signs, i.e. if the points are $$(x_1 , y_1)$$ and $$(x_2 , y_2)$$ then $$y_1 = -y_2 \neq 0$$.

Insert $$y = mx + 1$$ into the circle’s equation:

$$x^2 + (mx + 1)^2 + 3x = 0$$

$$\Rightarrow x^2 + m^2x^2 + 2mx + 1 + 3x = 0$$

$$\Rightarrow (1 + m^2)x^2 + (2m + 3)x + 1 = 0 \qquad -(1)$$

Equation (1) is quadratic in $$x$$ with roots $$x_1,\,x_2$$ corresponding to the two intersection points.

Their $$y$$-coordinates are $$y_1 = mx_1 + 1$$ and $$y_2 = mx_2 + 1$$. Since $$y_1 + y_2 = 0$$, we have

$$m(x_1 + x_2) + 2 = 0 \qquad -(2)$$

The sum of roots of (1) is

$$x_1 + x_2 = \frac{-(2m + 3)}{1 + m^2} \qquad -(3)$$

Substitute (3) into (2):

$$m\left(\frac{-(2m + 3)}{1 + m^2}\right) + 2 = 0$$

$$\Rightarrow \frac{-m(2m + 3)}{1 + m^2} = -2$$

Cross-multiply:

$$m(2m + 3) = 2(1 + m^2)$$

$$2m^2 + 3m = 2 + 2m^2$$

$$3m = 2$$

$$m = \frac{2}{3}$$

This relation can be written as $$3m - 2 = 0$$.

Hence, the required condition is satisfied only for

Option B which is: $$3m - 2 = 0$$