If in $∆ABC, a, b, c$are in AP and $p_1 , p_2 , p_3$ are the altitudes from the vertices $A, B, C$ respectively, then 

$\because \mathrm{\Delta }=\frac{1}{2}a{p}_1=\frac{1}{2}b{p}_2=\frac{1}{2}c{p}_3 (by figure)$

$\Rightarrow p_1=\frac{2\mathrm{\Delta }}{a},p_2=\frac{2\mathrm{\Delta }}{b},p_3=\frac{2\mathrm{\Delta }}{c}$

But given that $a, b, c$ are in AP.

Then, $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in HP

$\Rightarrow \frac{2\mathrm{\Delta }}{a},\frac{2\mathrm{\Delta }}{b},\frac{2\mathrm{\Delta }}{c}$ are also in HP.

$\Rightarrow p_1,p_2,p_3\text{ are in HP. }$   (option (b)] 

$\because p_1=\frac{2\mathrm{\Delta }}{a}, p_2=\frac{2\mathrm{\Delta }}{b}, p_3=\frac{2\mathrm{\Delta }}{\mathrm{C}}$

$\begin{array}{l}\Rightarrow p_1=\frac{2\mathrm{\Delta }}{2R\sin A},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{p}_2=\frac{2\mathrm{\Delta }}{2R\sin B},p_3=\frac{2\mathrm{\Delta }}{2R\sin C}\\ \Rightarrow \frac{1}{p_1}=\frac{R\sin A}{\mathrm{\Delta }},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{1}{p_2}=\frac{R\sin B}{\mathrm{\Delta }},\frac{1}{p_3}=\frac{R\sin C}{\mathrm{\Delta }}\end{array}$

$\because a,b,c$ are in AP.

Then $\sin A,\sin B,\sin C$ are also in AP.

$\Rightarrow 2\sin B=\sin A+\sin C ...\left(i\right)$

Now,  $\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}=\frac{R}{\mathrm{\Delta }}(\sin A+\sin B+\sin C)$

          $=\frac{R}{\mathrm{\Delta }}3\sin B [from Eq. (i)J$

         $\le \frac{3R}{\mathrm{\Delta }}$