Given,

5a–3b / a = 4a+b-2c / a+4b-2c = a+2b-3c / 4a-4c.

Prove that 6a= 4b= 3c.

Given (5a-3b) /a=(4a+b-2c) /(a+4b-2c) =(a+2b-3c) /(4a-4c)
Let each constant be equal to 'k'
=> (5a-3b) /a = k
=>  (5a-3b) = ka-------(1)
(4a+b-2c) /(a+4b-2c) = k
=>(4a+b-2c) = k(a+4b-2c)----(2)
(a+2b-3c) /(4a-4c) = k
=>(a+2b-3c) =k(4a-4c)-------(3)
Now, Adding (1)-(2)+(3), we get
2a-2b-c = k(4a-4b-2c)
=>2a-2b-c = 2k(2a-2b-c)
=> k  = 1/2
Thus, each ratio, k = 1/2.
Now substituting value of , back in equation(1), we get
b/a = 3/2---(*)
Substituting value of b/a = 3/2 and value of k =1/2 in any one of equations 
(2) or (3), we get
c/a = 2---(**)
From, (*) and (**), we get 6a =4b = 3c

 

Let 6 a = 4 b = 3 c = 12k (12 is the LCM of 6,4 and 3)

Hence,

a = 2k b = 3k  c = 4k.

 { 5 a - 3 b }/ { a } =  { 5 ( 2 k ) - 3 ( 3 k ) } /{ 2 k } =  { k }/ { 2 k } =  { 1 }/ { 2 }
 { 4 a + b - 2 c }/ { a + 4 b - 2 c } =  { 4 ( 2 k ) + ( 3 k ) - 2 ( 4 k ) }/ { 2 k + 4 ( 3 k ) - 2 ( 4 k )}= {3k}/{6k}= {1}/{2}
 { a + 2 b - 3 c }/ { 4 a - 4 c } =  { 2 k + 2 ( 3 k ) - 3 ( 4 k ) }/ { 4 ( 2 k ) - 4 ( 4 k ) }= {-4k}/{-8k}= {1}/{2}
Since all expressions are equal to {1}/{2}, therefore the assumption is correct.

Therefore, 6a = 4b = 3c