1. **State the problem:** Find the points of intersection between the line defined by the system $$\begin{cases} 2x - y + 2z - 12 = 0 \\ 2x - 4y - z + 6 = 0 \end{cases}$$ and the sphere $$x^2 + y^2 + z^2 - 2x + 2y + 4z - 43 = 0.$$\n\n2. **Rewrite the line equations to express variables in terms of a parameter:** From the first equation: $$2x - y + 2z = 12 \implies y = 2x + 2z - 12.$$\nSubstitute $y$ into the second equation: $$2x - 4(2x + 2z - 12) - z + 6 = 0.$$\nSimplify: $$2x - 8x - 8z + 48 - z + 6 = 0 \implies -6x - 9z + 54 = 0.$$\nRearranged: $$6x + 9z = 54 \implies 2x + 3z = 18.$$\nExpress $x$ in terms of $z$: $$x = \frac{18 - 3z}{2}.$$\nRecall $y = 2x + 2z - 12$, substitute $x$: $$y = 2\left(\frac{18 - 3z}{2}\right) + 2z - 12 = (18 - 3z) + 2z - 12 = 6 - z.$$\n\n3. **Parametric form of the line:** Let $z = t$, then $$x = \frac{18 - 3t}{2}, \quad y = 6 - t, \quad z = t.$$\n\n4. **Substitute parametric coordinates into the sphere equation:** $$\left(\frac{18 - 3t}{2}\right)^2 + (6 - t)^2 + t^2 - 2\left(\frac{18 - 3t}{2}\right) + 2(6 - t) + 4t - 43 = 0.$$\n\n5. **Simplify step-by-step:** $$\left(\frac{18 - 3t}{2}\right)^2 = \frac{(18 - 3t)^2}{4} = \frac{324 - 108t + 9t^2}{4} = 81 - 27t + \frac{9t^2}{4}.$$\n$$(6 - t)^2 = 36 - 12t + t^2.$$\n$$t^2 = t^2.$$\n$$-2\left(\frac{18 - 3t}{2}\right) = - (18 - 3t) = -18 + 3t.$$\n$$2(6 - t) = 12 - 2t.$$\n$$4t = 4t.$$\n\nSum all terms: $$81 - 27t + \frac{9t^2}{4} + 36 - 12t + t^2 + t^2 - 18 + 3t + 12 - 2t + 4t - 43 = 0.$$\nCombine like terms: Constants: $81 + 36 - 18 + 12 - 43 = 68.$\n$t$ terms: $-27t - 12t + 3t - 2t + 4t = -34t.$\n$t^2$ terms: $\frac{9t^2}{4} + t^2 + t^2 = \frac{9t^2}{4} + 2t^2 = \frac{9t^2}{4} + \frac{8t^2}{4} = \frac{17t^2}{4}.$\n\nEquation becomes: $$\frac{17t^2}{4} - 34t + 68 = 0.$$\nMultiply both sides by 4: $$17t^2 - 136t + 272 = 0.$$\n\n6. **Simplify by dividing entire equation by 17:** $$\cancel{17}t^2 - \cancel{17}8t + \cancel{17}16 = 0 \implies t^2 - 8t + 16 = 0.$$\n\n7. **Solve quadratic:** $$t^2 - 8t + 16 = 0.$$\nDiscriminant: $$\Delta = (-8)^2 - 4 \cdot 1 \cdot 16 = 64 - 64 = 0.$$\nOne real root: $$t = \frac{8}{2} = 4.$$\n\n8. **Find coordinates at $t=4$:** $$x = \frac{18 - 3(4)}{2} = \frac{18 - 12}{2} = \frac{6}{2} = 3,$$ $$y = 6 - 4 = 2,$$ $$z = 4.$$\n\n**Final answer:** The line intersects the sphere at the single point $$\boxed{(3, 2, 4)}.$$