Question : If $3a=4b=6c$ and $a+b+c=27\sqrt{29}$, then $\sqrt{a^{2}+b^{2}+c^{2}}$ is equal to:
Option 1: 87
Option 2: $3\sqrt{29}$
Option 3: 82
Option 4: 83
Correct Answer: 87
Solution :
Given: $3a=4b=6c$
⇒ $3a=6c$
⇒ $a=2c$
$4b=6c$
⇒ $b=\frac{3}{2}c$
Putting these values in the equation, we get,
$a+b+c=27\sqrt{29}$
⇒ $2c+\frac{3}{2}c+c=27\sqrt{29}$
⇒ $\frac{9}{2}c=27\sqrt{29}$
⇒ $c=6\sqrt{29}$
now we have to find $\sqrt{a^2+b^2+c^2}$, then:
$\sqrt{a^2+b^2+c^2}=\sqrt{(a+b+c)^2–2(ab+bc+ca)}$
Putting the values
⇒ $\sqrt{a^2+b^2+c^2}=\sqrt{(27\sqrt{29})^2–2(2c×\frac{3}{2}c+\frac{3}{2}c×c+c×2c)}$
⇒ $\sqrt{a^2+b^2+c^2}=\sqrt{(27\sqrt{29})^2–2(3c^2+\frac{3}{2}c^2+2c^2)}$
⇒ $\sqrt{a^2+b^2+c^2}=\sqrt{(729×29)–2×\frac{13}{2}c^2}$
⇒ $\sqrt{a^2+b^2+c^2}=\sqrt{(729×29)–13c^2}$
⇒ $\sqrt{a^2+b^2+c^2}=\sqrt{21141–13(6\sqrt{29})^2}$
⇒ $\sqrt{a^2+b^2+c^2}=\sqrt{21141–13(36×29)}$
⇒ $\sqrt{a^2+b^2+c^2}=\sqrt{21141–13572}$
⇒ $\sqrt{a^2+b^2+c^2}=\sqrt{7569}$
$\therefore \sqrt{a^2+b^2+c^2}=87$
Hence, the correct answer is 87.
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