If $$a^2 + 4b^2 + 49c^2 + 18 = 2(2b + 28c - a)$$, then the value of $$(3a + 2b + 7c)$$ is:

$$a^2 + 4b^2 + 49c^2 + 18 = 2(2b + 28c - a)$$

$$a^2 + 4b^2 + 49c^2 + 18 = 4b + 56c - 2a$$

$$a^2 + 4b^2 + 49c^2 + 18 - 4b - 56c + 2a = 0$$

$$a^2 + 2a + 4b^2 - 4b + 49c^2 - 56c + 18 = 0$$

$$(a^2 + 2a + 1) + (4b^2 - 4b+1) + (49c^2 - 56c + 16) = 0$$

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

$$(a + 1)^2 = 0$$

a = -1

$$(2b - 1)^2 = 0$$

2b - 1 = 0

$$b = \frac{1}{2}$$

$$(7c - 4)^2 = 0$$

7c - 4 = 0

$$c = \frac{4}{7}$$

value of $$(3a + 2b + 7c)$$ = $$(3\times(-1) + 2 \times \frac{1}{2} +7\times\frac{4}{7})$$
= (-3+1+4)

= 5-3

= 2