If x^{2}+4 y^{2}+3 z^{2}+\frac{19}{4} =2 \sqrt{3}(x+y+z), then the value of (x-4 y+3 z) is

SSC CHSL 5 Aug 2021 Shift 3 Paper

Section: Quantitative Aptitude

Question : 75 of 100

Marks: +1, -0

x^{2}+4 y^{2}+3 z^{2}+\;{19}/{4}

=2 √{3}(x+y+z)

, then the value of

(x-4 y+3 z)

Solution:

Given,

x^{2}+4 y^{2}+3 z^{2}+\;{19}/{4}

=2 √{3}(x+y+z)

⇒(x)^{2}+(2 y)^{2}+(√{3} z)^{2}+\;{19}/{4}

=2 √{3} x+2 √{3} y+2 √{3} z

⇒(x)^{2}-2 √{3} x+(√{3})^{2}+(2 y)^{2}

-2 √{3} y+(\;{√{3}}/{2})^{2}+(√{3} z)^{2}

-2 √{3} z+(1)^{2}=0

⇒(x-√{3})^{2}+(2 y-\;{√{3}}/{2})^{2}

+(√{3} z-1)^{2}=0

⇒(x-√{3})^{2}=0,(2 y-\;{√{3}}/{2})^{2}=0

(√{3} z-1)^{2}=0

⇒ x=√{3}, y \;\; =\;{√{3}}/{4}, z=\;{1}/{√{3}}

∴ x-4 y+3 z\;

\;\; =√{3}-4 × \;{√{3}}/{4}+3 × \;{1}/{√{3}}

\;\; =√{3}-√{3}+√{3}=√{3}

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