KCC's Quizzes AQQ288 about an amazing Equations System

Solving a,b,c based on k will give  a=sqrt(3k/2)  b==sqrt(2k/3)  c= sqrt(6k)

3k/2 and 2k/3 and 6k have common multiplied of 36..which is a squared value resulting in positive integer...

1)So k= 6...24....96...384..... multiplied by 4 which is also a square resulting in x2  
2) or k = 6....54....486.... multiplied by 9 which is also a square resulting in x3.
3) x4 is repeat of #1 above.
4) or k= 6...150....3750.... multiplied by 25 which is also a square resulting in x5...

It continues with all squared values 36, 49, 64, 81, 100, 121....

The answer is k(n)=6n2

CoPilot tells me: 

To find all values of ( k ) that yield positive integer solutions for the system:

Le paramètre k peut prendre les valeurs suivantes k=6.n.n avec n=1, 2, 3, ... Alors a=3.n, b=2.n et c=6.n sont >0.

Answer: k_n=6n^2, n positive integer

You can replace k=a*b in the second and the third equation and get 2a=c=3b. That means: a=(3/2)b. So b_n=2n, a_n=3n, c_n=6n. Since k=a*b, you get k_n=6n^2

The parameter k should take the folowing values: k=6.n.n with n=1, 2, 3, ... Then a=3.n, b=2.n and c=6.n are >0.

For n=1, k=6, a=3, b=2, c=6

For n=2, k=24, a=6, b=4, c=12

For n=3, k=54, a=9, b=6, c=18  etc...

Sorry, my first answer was curiously translated into French by my mobile.

k=6n²

^{n is an integer}

From the three equations relating a,b,and c:
 a = sqrt(1.5k)
 b = sqrt(k/1.5)
 c = sqrt(6k)

Plotting the three functions of k gives integer results at k=6.
k can then be scaled by an integer factor (1+n)² to give other sets of integer results.

k = 6(1+n)²  n = 0,1,2....

Where n is positive, 

a=3na equals 3 n

=3

b=2nb equals 2 n

 and

c=6nc equals 6 n

=6

a=3, b=2, c=6

K^2