First, the condition implies the given equation is solvable in real number. Therefore $$\Delta=(2a)^2 - 4\times(6-a)=4a^2 +4a -24 \ge 0 \implies a \ge 2\quad\text{or}\quad a\le -3$$ Let $x_1$ and $x_2$ be the two roots of this equation. Both of them are greater than 1 is equivalent to both of $(x_1-1)$ and $(x_2-1)$ are positive. This means that $$ \left\{ \begin{array}{cclclc} (x_1 - 1) + (x_2 - 1) &=& (x_1 + x_2) - 2 &=& -2a - 2 & > 0\\ (x_1 - 1)(x_2 - 1) &=& x_1x_2 - (x_1+x_2)+1 &=& 6-a+2a+1 & > 0 \end{array} \right. $$ Solving this system and also consider the permissible range for $a$ leads to $\boxed{-7 < a\le -3}$ Similarly, the condition for one root is greater than 1 and the other is less than 1 is equivalent to the product of $(x_1-1)$ and $(x_2-1)$ is negative. Or $$(x_1-1)(x_2-1) = x_1x_2 -(x_1+x_2)+1 = (6-a) + 2a +1 > 0 \implies \boxed{a < -7}$$