7. (10 points) The sum of the numbers on opposite faces of each cube is 7. For the three dice placed as shown in the figure, you can only see the numbers on seven faces. What is the sum of the numbers on all the faces that you cannot see from this figure? $\qquad$ .
7. (10 points) Each pair of opposite faces on a cubic die adds up to 7. In the arrangement of three dice shown, you can only see the numbers on seven faces, and the sum of the numbers on all the faces you cannot see is 41.
【Analysis】In this diagram, there are three cubic dice, numbered from top to bottom as (1)(2). Since the sum of the numbers on each pair of opposite faces of a cubic die is 7, the sum of the numbers on each die is composed of 3 sevens.

Therefore, the sum of the numbers on the faces of (1) that you cannot see is $7 \times 3-(4+3+2)=9$; the sum of the numbers on the faces of (2) that you cannot see is $7 \times 3-(5+1)=15$; the sum of the numbers on the faces of (3) that you cannot see is $7 \times 3-(3+1)=17$.

So, the sum of the numbers on all the faces you cannot see is $9+15+17=41$.
【Solution】Solution: (1) The sum of the numbers on the faces that you cannot see is: $7 \times 3-(4+3+2)=9$;
(2) The sum of the numbers on the faces that you cannot see is: $7 \times 3-(5+1)=15$;
(3) The sum of the numbers on the faces that you cannot see is: $7 \times 3-(3+1)=17$.

Therefore, the sum of the numbers on all the faces you cannot see is $9+15+17=41$.
The answer is: 41.