A common tangent T to the curves $$C_1: \frac{x^2}{4} + \frac{y^2}{9} = 1$$ and $$C_2: \frac{x^2}{42} - \frac{y^2}{143} = 1$$ does not pass through the fourth quadrant. If T touches $$C_1$$ at $$(x_1, y_1)$$ and $$C_2$$ at $$(x_2, y_2)$$, then $$|2x_1 + x_2|$$ is equal to _______.

We need to find $$|2x_1 + x_2|$$ where a common tangent T (not passing through the fourth quadrant) touches the ellipse $$C_1: \frac{x^2}{4} + \frac{y^2}{9} = 1$$ at $$(x_1, y_1)$$ and the hyperbola $$C_2: \frac{x^2}{42} - \frac{y^2}{143} = 1$$ at $$(x_2, y_2)$$.

For the ellipse $$\frac{x^2}{4} + \frac{y^2}{9} = 1$$ (here $$a^2 = 4, b^2 = 9$$), the equation of the tangent in slope-intercept form is $$y = mx \pm \sqrt{4m^2 + 9}$$.

Similarly, for the hyperbola $$\frac{x^2}{42} - \frac{y^2}{143} = 1$$ (here $$a^2 = 42, b^2 = 143$$), we have $$y = mx \pm \sqrt{42m^2 - 143}$$.

Since the tangent is common, the expressions under the square root must be equal, which yields $$4m^2 + 9 = 42m^2 - 143$$. Solving gives $$38m^2 = 152$$, so $$m^2 = 4$$ and $$m = \pm 2$$.

Substituting $$m^2 = 4$$ into $$c^2 = 4m^2 + 9$$ leads to $$c^2 = 25$$ and hence $$c = \pm 5$$.

Therefore, the four candidate tangents are $$y = 2x + 5$$, $$y = 2x - 5$$, $$y = -2x + 5$$, and $$y = -2x - 5$$.

To ensure the line does not enter the fourth quadrant ($$x > 0, y < 0$$), we require that $$y \ge 0$$ whenever $$x > 0$$. In this context, for $$y = 2x + 5$$, we have $$y > 5$$ for all $$x > 0$$, so the line always lies above the x-axis.

Next, recall that the point of tangency $$(x_1, y_1)$$ on the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ for a line $$y = mx + c$$ is given by $$\left(-\frac{a^2 m}{c}, \frac{b^2}{c}\right)$$.

With $$a^2 = 4$$, $$b^2 = 9$$, $$m = 2$$, and $$c = 5$$, it follows that $$x_1 = -\frac{4 \times 2}{5} = -\frac{8}{5}$$ and $$y_1 = \frac{9}{5}$$.

Verification: $$\frac{(-8/5)^2}{4} + \frac{(9/5)^2}{9} = \frac{64/25}{4} + \frac{81/25}{9} = \frac{16}{25} + \frac{9}{25} = 1$$, confirming the point lies on $$C_1$$.

Similarly, for the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$, the tangency point for $$y = mx + c$$ is given by $$\left(-\frac{a^2 m}{c}, -\frac{b^2}{c}\right)$$.

With $$a^2 = 42$$, $$b^2 = 143$$, $$m = 2$$, and $$c = 5$$, we find $$x_2 = -\frac{42 \times 2}{5} = -\frac{84}{5}$$ and $$y_2 = -\frac{143}{5}$$.

Verification: $$\frac{(-84/5)^2}{42} - \frac{(-143/5)^2}{143} = \frac{7056/25}{42} - \frac{20449/25}{143} = \frac{168}{25} - \frac{143}{25} = \frac{25}{25} = 1$$, confirming the point lies on $$C_2$$.

Additionally, substituting $$x_2$$ into $$y = 2x + 5$$ gives $$y_2 = 2\left(-\frac{84}{5}\right) + 5 = -\frac{168}{5} + \frac{25}{5} = -\frac{143}{5}$$, verifying consistency with the tangent line.

Finally, we compute $$2x_1 + x_2 = 2\left(-\frac{8}{5}\right) + \left(-\frac{84}{5}\right) = -\frac{16}{5} - \frac{84}{5} = -\frac{100}{5} = -20$$, and hence $$|2x_1 + x_2| = 20$$.

Therefore, the required value is $$\boxed{20}$$.