xyzlmk are real numbers such that x2y2z225 l2m2k236 and xlym

x,y,z,l,m,k are real numbers such that

x^2+y^2+z^2=25,

l^2+m^2+k^2=36, and

xl+ym+zk=30.

Evaluate (x+y+z)/(l+m+k)

x^2+y^2+z^2=25,

l^2+m^2+k^2=36

Add :

x^2 + l^2 + y^2 + m^2 + z^2 + k^2 = 25 + 36

(x + l)^2 + (y + m)^2 + (z + k)^2 - 2(xl + ym + zk) = 61

(x + l)^2 + (y + m)^2 + (z + k)^2 - 2(30) = 61

(x + l)^2 + (y + m)^2 + (z + k)^2 - 60 = 61

(x + l)^2 + (y + m)^2 + (z + k)^2 = 121

1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81 , 100

By looking at the above perfect squares, 4 + 36 + 81 = 121

So, x+l , y+m , z+k must be 2,6,9 in any order

From x^2 + y^2 + z^2 = 25
We can have 3 solution sets
5,0,0
0,5,0
0,0,5

From l^2 + m^2 + k^2 = 36
We can have 3 solution sets
6,0,0
0,6,0
0,0,6

It can only be :

x=5,y=0,z=0,l=6,m=0,n=0
OR
x=0,y=5,z=0,l=0,m=6,n=0
OR
x=0,y=0,z=5,l=0,m=0,k=6

In any of those cases, (x+y+z)/(l+m+k) would be :

(5+0+0)/(6+0+0) OR (0+5+0)/(0+6+0) OR (0+0+5)/(0+0+6)

All of the above result in :

5/6 ----> ANSWER