Introduction:
In a plane triangle ABC, we need to find out the maximum value of cosAcosBcosC.

Solution:
We know that in a triangle, the sum of all angles is 180 degrees, i.e., A+B+C=180.

Using the cosine rule, we can write cosA = (b^2+c^2-a^2)/(2bc), cosB = (a^2+c^2-b^2)/(2ac), and cosC = (a^2+b^2-c^2)/(2ab).

Substituting these values in the given expression, we get cosAcosBcosC = [(b^2+c^2-a^2)/(2bc)] [(a^2+c^2-b^2)/(2ac)] [(a^2+b^2-c^2)/(2ab)].

Simplifying this expression, we get cosAcosBcosC = [(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2)]/(8a^2b^2c^2).

Now, we need to find the maximum value of cosAcosBcosC.

AM-GM Inequality:
For any set of positive real numbers, the arithmetic mean is always greater than or equal to the geometric mean, i.e., (a+b+c)/3 >= (abc)^(1/3).

Using this inequality, we can write (a^2+b^2+c^2)/3 >= (a^2b^2c^2)^(1/3).

Multiplying both sides by 8/3, we get (8/3)(a^2+b^2+c^2)/3 >= (8/3)(a^2b^2c^2)^(1/3).

Now, substituting this inequality in the expression of cosAcosBcosC, we get cosAcosBcosC <=>

Therefore, the maximum value of cosAcosBcosC is (a^2+b^2+c^2)/9, which occurs when a=b=c, i.e., the triangle is an equilateral triangle.

Conclusion:
The maximum value of cosAcosBcosC in a plane triangle ABC is (a^2+b^2+c^2)/9, which occurs when the triangle is an equilateral triangle.