Capacitive Divider Confusion
For capacitors applies that the charge is conserved (as well as the energy), but the voltage will not be conserved, it will change.  

Giving the element reference designators, assuming "top plate" means the positive terminal and using the same definition for the voltage across C1 being \$V_{in}-V_{ref}\$ as OP does I come to this picture.



In the first situation, the total charge is:

$$ Q_{total} = (C_1+C_2+C_3)(V_{in}-V_{ref}) + (C_4+C_5+C_6)V_{in} $$
$$ Q_{total} = (4C)(V_{in}-V_{ref}) + (4C)V_{in} $$

In the second situation the total charge is:

$$ Q_{total} = (C_2+C_3)(V_{unknown}-V_{ref}) + (C_1+C_4+C_5+C_6)V_{unknown} $$
$$ Q_{total} = (2C)(V_{unknown}-V_{ref}) + (6C)V_{unknown} $$

As charge is conserved, you can find the total charge from the first equation and therefore be able to solve \$V_{unknown} = V_{DAC}\$.

Not sure whether this is a homework assignment, below is the spoiler:  


   $$(2C)(V_{unknown}-V_{ref}) + (6C)V_{unknown} = (4C)(V_{in}-V_{ref}) + (4C)V_{in} $$
 $$(8C)V_{unknown} = (8C)V_{in}-(2C)V_{ref} $$
 $$ V_{DAC} = V_{unknown} = V_{in}-\frac{1}{4}V_{ref} $$