If $a^2+b^2+49 c^2+18=2(b+28 c-a)$, then the value of $(2 a-b+7 c)$ is:

1

If $a^2+b^2+49 c^2+18=2(b+28 c-a)$

$(2 a-b+7 c)$ = ?

a² + b² + 49c² + 18 = 2(b - 28c - a)

= a² + b² + 49c² + 18 = 2b - 56c -2a

= a² + b² + 49c² + 18 - 2b + 56c + 2a = 0

= (a² + 2a + 1 ) + (b² - 2b + 1) + (7c)² + 56c + 16 = 0

= (a + 1)² + (b - 1)² + {(7c)² + 56c + 4²} = 0

= (a + 1)² + (b - 1)² + (7c + 4)² = 0 

Now,

a + 1 = 0

= a = -1

b -1 = 0

= b = 1

7c + 4 = 0

= c = \(\frac{4}{7}\)

$(2 a-b+7 c)$ = (2 × -1 - 1 +7  × \(\frac{4}{7}\))  = 1