Let $CK$ meet $AB$ at $F$ then from Seva teorem $\frac{AE}{EC}\frac{CD}{DB}\frac{BF}{FA}=1 \Longrightarrow AE=BD \Longrightarrow\frac{CD}{EC}=\frac{FA}{FB}\Longrightarrow \frac{CD+CE}{EC}=\frac{AB}{FB}$ $\Longrightarrow EC=FB$ and $AF=CD$ So $AF=CD=p-b, BF=CE=p-a, BD=AE=p-c$ Let $A_1,B_1,C_1$ be the feet of altitudes from $A,B,C$, respectively.Let $K_1,K_2$ be feet of perpendiculars from $K$to the side $BC$ and to the altitude $AA_1$ and let $S,T$ and $R$ be the feet of perpendiculars from $I$ and $O$ to the side $BC$, and foot of altitude $O$ to $IS$, respectively. From the Seva teorem we get $\frac{AD}{KD}=\frac{p}{p-a}=\frac{AA_1}{KK_1} \Longrightarrow KK_1=\frac{p-a}{p}AA_1$ $|HK_2|=|HA_1-KK_1|=|AA_1-AH-AA_1\frac{p-a}{p}|=|\frac{a}{p}AA_1-AH|=|\frac{2S}{p}-AH|=|2r-2OT|$ $\Longrightarrow HK_2=2IR$ $(1)$ $KK_2=A_1K_1=A_1D-K_1D=A_1D-A_1D \frac{KD}{AD}=A_1D-A_1D\frac{p-a}{p}=\frac{a}{p}A_1D=\frac{a}{p}(CA_1-CD)=\frac{a}{p}(\frac{a^2+b^2-c^2}{2a}-(p-b))=b-c$ $\Longrightarrow KK_2=2(p-c-\frac{a}{2})=2ST$ $\Longrightarrow KK_2=2OR$$(2)$ Using $(1)$,$(2)$ and $\angle KK_2H=\angle ORI=\frac{\pi}{2}$ we get $\bigtriangleup KK_2H \sim \bigtriangleup ORI$ with ratio $2$. So we get $KH \Vert OI$ and $KH=2OI$

We have $(p-a)\overrightarrow{KA}+(p-b)\overrightarrow{KB}+(p-c)\overrightarrow{KC}=\overrightarrow{0}$, $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OH}$, $a\overrightarrow{IA}+b\overrightarrow{IB}+c\overrightarrow{IC}=\overrightarrow{0}$. Now, it is easy !

It's very easy problem.We have $ K $ is Nagel point of the triangle $ ABC $ and by Nagel line theorem $ K,I,G $ are collinear and $ GK=2IG $, on the other hand $ H,G,O $ are collinear ( Euler line ) and $ HO=2OG $ ,so the triangle $ IGO $ and the triangle $ HGK $ similar, so $ IO $ parallel to $ HK $ and $ HK=2IO $ .

Where $ G $ is centroid of the trianle $ ABC $.