SSC CGL 20 Aug 2021 Shift 1 Paper

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Question : 60 of 100

Marks: +1, -0

If

a^{2}+b^{2}+c^{2}+216

=12(a+b-2 c)

, then

√{a b-b c-c a}

is

$3 √{5}$
$8 √{5}$
$6 √{5}$
$4 √{5}$

Solution:

Given,

a^{2}+b^{2}+c^{2}+216

⇒ a^{2}+b^{2}+c^{2}+216(a+b-2 c)

\;\;\; =12 a+12 b-24 c

⇒ a^{2}-12 a+b^{2}-12 b

\;\;\; +c^{2}+24 c+36+36+144=0

⇒(a^{2}+(6)^{2}-2 × 6 × a)+(b^{2}+6^{2}

\;\; -2 × 6 × b)+(c^{2}+(12)^{2}+2 × 12 × c)

=0

⇒(a-6)^{2}+(b-6)^{2}+(c+12)^{2}=0

⇒ a=6, b=6, c=-12

∴ √{a b-b c-c a}

\;\; =√{[6 × 6-6 ×(-12)-(-12 × 6)]}

\;\; =√{36+72+72}=√{180}

\;\; =√{6 × 6 × 5}=6 √{5}

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