Since the triangle sides are diameters of the circles $k_{1}, k_{2}, k_{3},$ their pairwise radical axes are the triangle altitudes and their radical center K is the triangle orthocenter. Let $h_{a}= AA',\ h_{b}= BB',\ h_{c}= CC'$ be the A-, B-, C-altitudes of $\triangle ABC$ and let $h_{d}= DA',\ h_{e}= EB',\ h_{f}= FC'$ be the D-, E-, F-altitudes of $\triangle BCD,\ \triangle CAE,\ \triangle ABF.$ $h_{a}= b \sin C = c \sin B,\ h_{b}= c \sin A = a \sin C,\ h_{c}= a \sin B = b \sin A$ $h_{a}^{2}= bc \sin B \sin C,\ h_{b}^{2}= ca \sin C \sin A,\ h_{c}^{2}= ab \sin A \sin B$ Powers of A', B', C' to the circles $k_{1}, k_{2}, k_{3}$ are $h_{d}^{2}= BA' \cdot CA' = bc \cos B \cos C$ $h_{e}^{2}= CB' \cdot AB' = ca \cos C \cos A$ $h_{f}^{2}= AC' \cdot BC' = ab \cos A \cos B$ $\frac{x^{2}+y^{2}+z^{2}}{u^{2}}= \frac{h_{d}^{2}}{h_{a}^{2}}+\frac{h_{e}^{2}}{h_{b}^{2}}+\frac{h_{f}^{2}}{h_{c}^{2}}=$ $= \cot B \cot C+\cot C \cot A+\cot A \cot B = 1$ for $\angle A+\angle B+\angle C = 180^\circ.$

Lemma 1 (well-known). Let $ABC$ be a triangle $ABC$ with the orthocenter $H\ .$ Denote the points $D\in AH\cap BC$ and the reflection $D_{1}$ of the point $H$ w.r.t. the line $BC\ .$ Then the point $D_{1}$ belongs to the circumcircle of the triangle $ABC$ and exists the relation $\boxed{\ DB\cdot DC=AD\cdot HD\ }\ .$ Lemma 2. Denote the area $[XYZ]$ of the triangle $XYZ\ .$ Let $ABC$ be an acute triangle with the orthocenter $H$. Denote the point $D\in AH\cap BC$ and $D'\in (AH)$ so that $D'B\perp D'C\ .$ Then $\boxed{\ [BD'C]^{2}=[BAC]\cdot [BHC]\ }\ .$ Proof. $\frac{[BD'C]^{2}}{[BAC]\cdot [BHC]}=$ $\frac{[BD'C]}{[BAC]}\cdot\frac{[BD'C]}{[BHC]}=$ $\frac{D'D}{AD}\cdot \frac{D'D}{HD}=\frac{DB\cdot DC}{AD\cdot HD}=1$ $\Longrightarrow$ $[BD'C]^{2}=[BAC]\cdot [BHC]\ .$ Remark. Analogously define the pairs of points $\{\ E\in BH\cap CA\ ;\ E'\in (BH)\ ,\ E'C\perp E'A\ \}\ ,$ $\{\ F\in CH\cap AB\ ;\ F'\in (CH)\ ,\ F'A\perp F'B\ \}$ $\Longrightarrow$ $\boxed{\ \sum_{\mathrm{cyc}}[BD'C]^{2}=[ABC]\cdot \sum_{\mathrm{cyc}}[BHC]=[ABC]^{2}\ }\ .$

Radical center is orthocenter of $\triangle ABC$. Let the altitudes drawn from $A$, $B$ and $C$ be $H_A$, $H_B$ and $H_C$, respectively. Then using the well known property $AH_A\cdot HH_A=BH_A\cdot CH_A$ (Reflect $H$ across $BC$ and it is PoP in $(ABC)$). Since $BD\perp CD$, $ AE\perp CE$ and $AF\perp BF$, we have $$[DBC]^2=\dfrac{BD^2}{CD^2}{4}=\dfrac{BH_A\cdot BC}{2}\cdot \dfrac{CH_A \cdot BC}{2}$$Using the property $$\boxed{[DBC]^2=\dfrac{BH_A\cdot BC}{2}\cdot \dfrac{CH_A \cdot BC}{2}=\dfrac{AH_A\cdot BC}{2}\cdot \dfrac{HH_C\cdot BC}{2}=[ABC]\cdot [HBC]}$$Thus, $$[DBC]^2=[ABC][HBC]$$$$[ECA]^2=[ABC][HCA]$$$$[FAB]^2=[ABC][HAB]$$After summing up, the result follows.