Given two points are (3sin θ, 0, 0) and (4cos θ, 0, 0)

Now distance = √{(4cos θ - 3sin θ)² + (0 - 0)² + (0 - 0)² }

=> distance = √{(4cos θ - 3sin θ)² }

=> distance = 4cos θ - 3sin θ  ................1

Now, maximum value of 4cos θ - 3sin θ = √{(4² + (-3)² }

                                                              = √(16 + 9)

                                                              = √25

                                                              = 5 

From equation 1, we get

distance = 5

So, the maximum distance between points (3sin θ, 0, 0) and (4cos θ, 0, 0) is 5