According to the principle of similarity ratio and the requirements of the topic,

(x + 3y)/(6 + 3×4)=(3y - 2z)/(3×4 - 2×3)

Therefore, (x+3y)/(3y-2z) = (6+3× 4)/(3× 4-2× 3) =18/6 = 3.

The second question:

Because the area of the parallelogram ABCD is 10, the height of the parallelogram is 10/AB = 10/5 = 2.

Because of the similarity, AB/2 = A'B'/h, it can be seen that the height of quadrilateral A'B'C'D' h = 3.

So the area of the parallelogram A'B'C'D' is A'B' × h = 6 × 3 = 18.

The third question:

There are two possibilities to make the triangle ACB and CBD similar:

1. If the AC side corresponds to the BD side, then the two triangles are congruent and similar, then BD = AC = B.

2. If the AC side is opposite to the BC side, then AC/BC = BC/BD and BD = A× A/B..