Here it is worked out using algebra:

Let N be the answer.  Then there exist positive integers A, B, and C
such that:

N = 3A + 1 = 4B + 2 = 5C + 3

3A + 1 = 4B + 2

3A - 4B = 1

Principle  P:
This is the principle of

1. Writing the absolute value  of 
each integer or coefficient in an equation in terms 
of its nearest multiple of the coefficient in absolute 
value that occurs in the equation,considering 0 as a 
multiple of every integer,
2. Removing parentheses
3. Dividing every term by that least coefficient in
absolute value.
4. Isolating all fractional 
terms.
5. Setting both sides equal to a new positive integer 
variable:

Using principle P on

3A - 4B = 1

3A - (3 + 1)B = 1

3A - 3B - B = 1

A - B - B/3 = 1/3

A - B = 1/3 + B/3

Let D = A - B = 1/3 + B/3

D = 1/3 + B/3

3D = 1 + B

(1)    B = 3D - 1

D = A - B 
D = A - (3D - 1)
D = A - 3D + 1
4D - 1 = A

(2)   A = 4D - 1

       4B + 2 = 5C + 3
4(3D - 1) + 2 = 5C + 3
  12D - 4 + 2 = 5C + 3
      12D - 2 = 5C + 3
     12D - 5C = 5

Use principle P on that equation:

 (10+2)D - 5C = 5
10D + 2D - 5C = 5
2D + 2D/5 - C = 1
2D/5 = C - 2D + 1

2D/5 = C - 2D + 1 = E

C - 2D + 1 = E

(3)   C = E + 2D - 1


2D/5 = E
2D = 5E

Use the principle P again:

2D = (4+1)E
2D = 4E + E
D = 2E + E/2
D - 2E = E/2

E/2 = D - 2E = F

E/2 = F,  D - 2E = F

E = 2F

D - 2(2F) = F
D - 4F = F
(4) D = 5F

From (3) 

C = E + 2D - 1
C = 2F + 2(5F) - 1
C = 2F + 10F - 1
C = 12F - 1

(5)  C = 12F - 1

From (1)    

B = 3D - 1
B = 3(5F) - 1
(6)  B = 15F - 1

D = A - B
5F = A - (15F - 1)
5F = A - 15F + 1

20F - 1 = A

A = 20F - 1
B = 15F - 1
C = 12F - 1

Therefore the smallest values of A, B, and C
will be when F=11

A = 20F - 1 = 20 - 1 = 19
B = 15F - 1 = 15 - 1 = 14
C = 7F - 1 = 12 - 1 = 11

N = 3A + 1 = 4B + 2 = 5C + 3

N = 3(19) + 1 = 4(14) + 2 = 5(11) + 3

N = 58 = 58 = 58

Edwin