SSC CHSL 2021 PYQ — Question 68

The correct answer is Option 1 i.e. ${{2} \over 21}$

Given:

${{9a^2+4b^2+49c^2+18=2(2b+28c-3a)}}$

Explanation:

According to the given question,

${{9a^2+4b^2+49c^2+18= 4b +56c- 6a}}$

⇒ (9a² + 6a + 1) + (4b² - 4b + 1) + (49c² - 56c + 16) = 0

⇒ (3a + 1)² + (2b - 1)² + ( 7c -4)² = 0

⇒ 3a + 1 = 0 ; 2b - 1 = 0 ; 7c - 4 = 0

⇒ a = $- {1\over 3}$ ; b = ½ ; c = $4 \over 7$

Now, required value = a + 2b - c = $-{1 \over 3} + 1 - {4 \over 7}$ = ${{2} \over 21}$

Hence the required answer is ${{2} \over 21}$.