If in a triangle ABC, a,b,c are in A.P. and $p_{1}, p_{2,} p_{3}$ are the altitudes from the vertices A, B, C Respectively then

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$\sin A, \sin B, \sin C$ are in A.P.

$\sin A, \sin B, \sin C$ are in H.P.

$p_{1} + p_{2} + p_{3} \leq \frac{3 R}{Δ}$

$\frac{1}{p_{1}} + \frac{1}{p_{2}} + \frac{1}{p_{3}} \leq \frac{3 R}{Δ}$

answer is A, D.

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Detailed Solution

$Δ = \frac{1}{2} a p_{1} = \frac{1}{2} b p_{2} = \frac{1}{2} c p_{3}$
$\Rightarrow p_{1} = \frac{2 Δ}{a}, p_{2} = \frac{2 Δ}{b}, p_{3} = \frac{2 Δ}{c}$
$p_{1}, p_{2}, p_{3}$ are in H.P.
Now $p_{1} = \frac{2 Δ}{2 R \sin A}, p_{2} = \frac{2 Δ}{2 R \sin B}, p_{3} = \frac{2 Δ}{2 R \sin C}$
$\Rightarrow \frac{1}{p_{1}} = \frac{R \sin A}{Δ}, \frac{1}{p_{2}} = \frac{R \sin B}{Δ}, \frac{1}{p_{3}} = \frac{R \sin C}{Δ}$
$\Rightarrow \sin A, \sin B, \sin C$ are in A.P.
$\Rightarrow \sin A + \sin C = 2 \sin B$
$∴ \frac{1}{p_{1}} + \frac{1}{p_{2}} + \frac{1}{p_{3}} = \frac{R}{Δ} (\sin A + \sin B + \sin C)$
= $\frac{3 R}{Δ} \sin B \leq \frac{3 R}{Δ}$