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If in a triangle $A B C, r_{1} = 2 r_{2} = 3 r_{3}$ , then the perimeter of the triangle is

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$3 a$

$4 a$

$3 b$

$3 c$

answer is C.

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Detailed Solution

$r_{1} = 2 r_{2} = 3 r_{3}$

$\frac{Δ}{s - a} = \frac{2 Δ}{s - b} = \frac{3 Δ}{s - c}$

$\frac{s - a}{s - b} = \frac{1}{2} \to (1)$

$\frac{s - c}{s - b} = \frac{3}{2} \to (2)$

$(1) + (2)$

$\frac{s - a + s - c}{s - b} = \frac{1}{2} + \frac{3}{2}$

$\frac{b}{s - b} = 2$

$b = 2 s - 2 b$

$2 s = 3 b$

$a + b + c = 3 b$

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