If x2 + 4y2 + 3z2 + 19/4 = 2 sqrt 3 (x+y+z) find (x-4y+3z).

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

Now we can say

x = \(\frac{Coefficient\; of\; x}{Coefficient\; of\; x^2}\) = \(\sqrt {3}\)

y = \(\frac{Coefficient\; of\; y}{Coefficient\; of\; y^2}\) = \(\frac{\sqrt {3}}{4}\)

z = \(\frac{Coefficient\; of\; z}{Coefficient\; of\; z^2}\) = \(\frac{\sqrt {3}}{3}\) = \(\frac{1}{\sqrt {3}}\)

Put in x - 4y + 3z

⇒ \(\sqrt {3}\) - 4 × \(\frac{\sqrt {3}}{4}\) + 3 × \(\frac{1}{\sqrt {3}}\)

⇒  \(\sqrt {3}\)