Let `xyz=105` where `x, y,z in N`. Then number of ordered triplets (x, y,z) satisfying the given equation is

xyz=24 , x,y,z in N. Then find the number of order pairs (x,y,z)

If x, y, z are natural numbers such that cot^(-1)x+cot^(-1)y=cot^(-1)z then the number of ordered triplets (x, y, z) that satisfy the equation is

If x,y,z are natural number such that cot^(-1)x+cot^(-1)y=cot^(-1)z then the number of ordered triplets (x,y,z) that satisfy the equation is

If x,y,z are natural numbers such that cot ^(-1)x+cot^(-1)y=cot^(-1)z then the number of ordered triplets (x,y,z) that satisfy the equation is 0 (b) 1 (c) 2 (d) Infinite solutions

If x , y , z are natural numbers such that cot^(-1)x+cot^(-1)y=cot^(-1)z then the number of ordered triplets (x , y , z) that satisfy the equation is 0 (b) 1 (c) 2 (d) Infinite solutions

If x, y, z are natural numbers such that cot^(-1) x + cot^(-1)y= cot^(-1) z then the number of ordered triplets (x, y, z) that satisfy the equation is a)0 b)1 c)2 d)infinite solutions

If x , y , z are natural numbers such that cot^(-1)x+cot^(-1)y=cot^(-1)z then the number of ordered triplets (x , y , z) that satisfy the equation is 0 (b) 1 (c) 2 (d) Infinite solutions

If x , y , z are natural numbers such that cot^(-1)x+cot^(-1)y=cot^(-1)z then the number of ordered triplets (x , y , z) that satisfy the equation is (a)0 (b) 1 (c) 2 (d) Infinite solutions

The number of ordered triplets, positive integers which are solutions of the equation x+y+z=100 is:

The number of ordered triplets, positive integers which are solutions of the equation x+y+z=100 is: