Let the sides of the triangle be a, b, and c.
The altitudes are given in the ratio of 1:2:3.
So we can let x be such that the altitudes have lengths x,2x, and 3x.

Let x be the length of the altitude drawn to side a.
Let 2x be the length of the altitude drawn to side b.
Let 3x be the length of the altitude drawn to side c.

Now we use the usual formula for the area of a triangle, which is

Area of a triangle = ·(any side of triangle)·(altitude drawn to that side)

Therefore

Area of the triangle = a·x = b·2x = ·3x

a·x = b·2x = c·3x

Multiply by 2

a·x = 2b·x = c·3x

ax = 2bx = 3cx

Divide by x

a = 2b = 3c

We are given that the perimeter is 36, so

P = a + b + c = 36

So we have the system

 a = 2b
 a = 3c
 a + b + c = 36

Solve the first eq. for b:  b = 
Solve the second eq. for c:  c = 

  a +  +  = 36
 6a + 3a + 2a = 216
          11a = 216
            a = 

Then b =  = a = · = 

And c =  = a = · = 

So the sides would have to be a=, b=, c=

However, the sum of any two sides of a triangle must be greater than
the third side.

So those values are not possible sides for a triangle because b+c < a

So there is no possible solution.  No triangle can have its altitudes in
the ratio 1:2:3.  

Edwin