$a, b, c$ are in A.P. and the points $A(a, 1), B(b, 2)$ and $C(c, 3)$ are such that $(OA)^{2}, (OB)^{2}$ and $(OC)^{2}$ are also in A.P; $O$ being the origin, then

a,b,c are in A.P

then

$2b=a+c$

now

${OA}^{2},{OB}^{2},{OC}^{2}$ are also in A.P

$\therefore 2{OB}^{2}={OA}^{2}+{OC}^{2}$

Now using distance formula,

${OA}^{2}={a}^{2}+1$

${OB}^{2}={b}^{2}+4$

${OA}^{2}={c}^{2}+9$

$2({b}^{2}+4)={a}^{2}+1+{c}^{2}+9$

On solving, we get

$2{b}^{2}={a}^{2}+{c}^{2}+2$

${a}^{2}+{c}^{2}=2{b}^{2}-2$---------(A)

Again

$2{b}^{2}={a}^{2}+{c}^{2}+2$

$2{b}^{2}={(a+c)}^{2}-2ac+2$ (since $a+c=2b$)

$2{b}^{2}={(2b)}^{2}-2ac+2$

on solving we get

$ac={b}^{2}+1$ ---- (B)

Now

$2{b}^{2}={a}^{2}+{c}^{2}+2$

$2{b}^{2}={(a+c)}^{2}-2ac+2$

$2{b}^{2}={(a+c)}^{2}-2({b}^{2}+1)+2$ by putting the value of ac obtained in B

on solving, we get

${(a+c)}^{2}=4{b}^{2}$ --- (C)

$2b=a+c$

adding b to both sides then we get

$3b=a+b+c$