If the system of equations
$$kx + y + 2z = 1$$
$$3x - y - 2z = 2$$
$$-2x - 2y - 4z = 3$$
has infinitely many solutions, then $$k$$ is equal to ______.

For the system $$kx + y + 2z = 1$$, $$3x - y - 2z = 2$$, $$-2x - 2y - 4z = 3$$ to have infinitely many solutions, the third equation must be a linear combination of the first two, in both the coefficients and the right-hand side.

Let $$R_3 = \alpha R_1 + \beta R_2$$. Comparing coefficients of $$y$$: $$-2 = \alpha(1) + \beta(-1) = \alpha - \beta$$. Comparing coefficients of $$z$$: $$-4 = \alpha(2) + \beta(-2) = 2(\alpha - \beta)$$, which gives $$\alpha - \beta = -2$$, consistent with the $$y$$-equation.

So $$\alpha = \beta - 2$$. Comparing coefficients of $$x$$: $$-2 = \alpha k + 3\beta = (\beta - 2)k + 3\beta = \beta(k + 3) - 2k$$, giving $$\beta(k + 3) = 2k - 2$$, hence $$\beta = \frac{2(k - 1)}{k + 3}$$.

Comparing the right-hand sides: $$3 = \alpha(1) + \beta(2) = (\beta - 2) + 2\beta = 3\beta - 2$$, which gives $$3\beta = 5$$, so $$\beta = \frac{5}{3}$$.

Setting the two expressions for $$\beta$$ equal: $$\frac{2(k-1)}{k+3} = \frac{5}{3}$$. Cross-multiplying: $$6(k - 1) = 5(k + 3)$$, so $$6k - 6 = 5k + 15$$, giving $$k = 21$$.

Therefore, $$k = 21$$.