If x^{2}+9 y^{2}+4 z^{2}=12 (x-2 y+2 z)-88, then the value of (x-3 y+z) is

SSC CGL 17 Aug 2021 Shift 3 Paper

Question : 64 of 100

Marks: +1, -0

x^{2}+9 y^{2}+4 z^{2}=12

(x-2 y+2 z)-88

, then the value of

(x-3 y+z)

Solution:

Given,

x^{2}+9 y^{2}+4 z^{2}=12(x-2 y+2 z)-88

⇒(x^{2}-12 x+36)+[(3 y)^{2}+24 y+16]

+[(2 z)^{2}-24 z+36]=0

⇒(x-6)^{2}+(3 y+4)^{2}+(2 z-6)^{2}=0

⇒ x=6, y=\;{-4}/{3}, z=3

∴ (x-3 y+z)=6-3 ×(-\;{4}/{3})+3

\;\; =6+4+3=13

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