In the triangle ABC, it is known that the opposite sides of angles a, B and C are a, B and C respectively, and (a + B + C) (B + C-A) = 3bC 3. If a = root 3, find the area of triangle ABC and the maximum value of triangle s

(a + B + C) (B + C-A) = 3bC [K ^ 2] is the square of K. similarly, the multiplication sign is dot multiplication·
（b+c+a)(b+c-a)=3bc
(b+c)^2-a^2=3bc
b^2+c^2-a^2=bc
Then divide the two sides by 2BC
What theorem can we get cos a = 1 / 2
Then. Then I don't know. After the college entrance examination, I will die. I forget all about it. Sorry
The following should be solved by using the relationship between edges and corners. Cosa can be transformed into other forms. If we multiply the square of both sides of 2B = 3C by B or C at the same time, we need to eliminate one of B or C in short. Then we can use cosine theorem and sine theorem several times more

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