SSC CHSL 2021 PYQ — Question 53

The correct answer is option 3 i.e. $\sqrt{3}$

(a - b)² = a² - 2ab + b²

$If A^2 + B^2 + C^2 = 0 => A = B = C = 0$

x² + 4y² + 3z² + $\frac{19}{4}=2√{3}(x+y+z)$

⇒ (x² - 2√3x + 3) + (4y² - 2√3y + (3/4)) + (3z² - 2√3z + 1) = 0

⇒ (x - √3)² + (2y - (√3/2))² + (√3z - 1)² = 0

So,

(x - √3)² = 0

⇒ x = √3

(2y - (√3/2))² = 0

⇒ y = √3/4

(√3z - 1)² = 0

⇒ z = 1/√3

Now,

x - 4y + 3z

⇒ √3 - 4 × (√3/4) + 3 × (1/√3)

⇒ √3 - √3 + √3

⇒ √3