Find all values of the real number $a$ so that the four complex roots of
\[z^4 - 6z^3 + 11az^2 - 3(2a^2 + 3a - 3) z + 1 = 0\]form the vertices of a parallelogram in the complex plane.  Enter all the values, separated by commas.
By Vieta's formulas, the average of the sum of the roots is $\frac{6}{4} = \frac{3}{2},$ which corresponds to the center of the parallelogram.  So, to shift the center of the parallelogram to the origin, let $w = z - \frac{3}{2}.$  Then $z = w + \frac{3}{2},$ so
\[\left( w + \frac{3}{2} \right)^4 - 6 \left( w + \frac{3}{2} \right)^3 + 11a \left( w + \frac{3}{2} \right)^2 - 3(2a^2 + 3a - 3) \left( w + \frac{3}{2} \right) + 1 = 0.\]Hence,
\[(2w + 3)^4 - 2 \cdot 6 (2w + 3)^3 + 4 \cdot 11a (2w + 3)^2 - 8 \cdot 3(2a^2 + 3a - 3)(2w + 3) + 16 = 0.\]Expanding, we get
\[16w^4 + (176a - 216) w^2 + (-96a^2 + 384a - 288) w - 144a^2 + 180a - 11 = 0.\]The roots of this equation will form a parallelogram centered at the origin, which means they are of the form $w_1,$ $-w_1,$ $w_2,$ $-w_2.$  Thus, we can also write the equation as
\[(w - w_1)(w + w_1)(w - w_2)(w + w_2) = (w^2 - w_1^2)(w^2 - w_2^2) = 0.\]Note that the coefficient of $w$ will be 0, so
\[-96a^2 + 384a - 288 = 0.\]This equation factors as $-96(a - 1)(a - 3) = 0,$ so $a = 1$ or $a = 3.$

For $a = 1,$ the equation becomes
\[16w^4 - 40w^2 + 25 = (4w^2 - 5)^2 = 0,\]which has two double roots.

For $a = 3,$ the given equation becomes
\[w^4 + 312w^2 - 767 = 0.\]The roots of $x^2 + 312x - 767 = 0$ are real, and one is positive and the other is negative.  This mean that two of the roots of $w^4 + 312w^2 - 767 = 0$ are real (and negatives of each other), and the other two are imaginary (and negatives of each other), so they form a parallelogram.

Thus, the only such value of $a$ is $\boxed{3}.$