Throughout the solutions, $\Varangle(p, q)$ will denote the directed angle between lines $p$ and $q$, taken modulo $180^{\circ}$.
Let the vertices of $\Delta_{i}$ be $D_{i}, E_{i}, F_{i}$, such that lines $E_{i}F_{i}$, $F_{i}D_{i}$ and $D_{i}E_{i}$ are the perpendiculars through $X, Y$ and $Z$, respectively, and denote the circumcircle of $\Delta_{i}$ by $\omega_{i}$.
In triangles $D_{1}Y_{1}Z_{1}$ and $D_{2}Y_{2}Z_{2}$ we have $Y_{1}Z_{1} \parallel Y_{2}Z_{2}$ because they are parts of $\ell_{1}$ and $\ell_{2}$. Moreover, $D_{1}Y_{1} \parallel D_{2}Y_{2}$ are perpendicular to $AC$ and $D_{1}Z_{1} \parallel D_{2}Z_{2}$ are perpendicular to $AB$, so the two triangles are homothetic and their homothetic centre is $Y_{1}Y_{2} \cap Z_{1}Z_{2} = A$. Hence, line $D_{1}D_{2}$ passes through $A$. Analogously, line $E_{1}E_{2}$ passes through $B$ and $F_{1}F_{2}$ passes through $C$.

The corresponding sides of $\Delta_{1}$ and $\Delta_{2}$ are parallel, because they are perpendicular to the respective sides of triangle $ABC$. Hence, $\Delta_{1}$ and $\Delta_{2}$ are either homothetic, or they can be translated to each other. Using that $B, X_{2}, Z_{2}$ and $E_{2}$ are concyclic, $C, X_{2}, Y_{2}$ and $F_{2}$ are concyclic, $Z_{2}E_{2} \perp AB$ and $Y_{2}, F_{2} \perp AC$ we can calculate
$$\begin{align*} \Varangle\left(E_{1}E_{2}, F_{1}F_{2}\right) &= \Varangle\left(E_{1}E_{2}, X_{1}X_{2}\right) + \Varangle\left(X_{1}X_{2}, F_{1}F_{2}\right) = \Varangle\left(BE_{2}, BX_{2}\right) + \Varangle\left(CX_{2}, CF_{2}\right) \\ &= \Varangle\left(Z_{2}E_{2}, Z_{2}X_{2}\right) + \Varangle\left(Y_{2}X_{2}, Y_{2}F_{2}\right) = \Varangle\left(Z_{2}E_{2}, \ell_{2}\right) + \Varangle\left(\ell_{2}, Y_{2}F_{2}\right) \\ &= \Varangle\left(Z_{2}E_{2}, Y_{2}F_{2}\right) = \Varangle(AB, AC) \not\equiv 0, \tag{1} \end{align*}$$
and conclude that lines $E_{1}E_{2}$ and $F_{1}F_{2}$ are not parallel. Hence, $\Delta_{1}$ and $\Delta_{2}$ are homothetic; the lines $D_{1}D_{2}$, $E_{1}E_{2}$, and $F_{1}F_{2}$ are concurrent at the homothetic centre of the two triangles. Denote this homothetic centre by $H$.
For $i=1,2$, using (1), and that $A, Y_{i}, Z_{i}$ and $D_{i}$ are concyclic,
$$\begin{aligned} \Varangle\left(HE_{i}, HF_{i}\right) &= \Varangle\left(E_{1}E_{2}, F_{1}F_{2}\right) = \Varangle(AB, AC) \\ &= \Varangle\left(AZ_{i}, AY_{i}\right) = \Varangle\left(D_{i}Z_{i}, D_{i}Y_{i}\right) = \Varangle\left(D_{i}E_{i}, D_{i}F_{i}\right), \end{aligned}$$
so $H$ lies on circle $\omega_{i}$.
The same homothety that maps $\Delta_{1}$ to $\Delta_{2}$, sends $\omega_{1}$ to $\omega_{2}$ as well. Point $H$, that is the centre of the homothety, is a common point of the two circles, That finishes proving that $\omega_{1}$ and $\omega_{2}$ are tangent to each other.

As in the first solution, let the vertices of $\Delta_{i}$ be $D_{i}, E_{i}, F_{i}$, such that $E_{i}F_{i}$, $F_{i}D_{i}$ and $D_{i}E_{i}$ are the perpendiculars through $X_{i}, Y_{i}$ and $Z_{i}$, respectively. In the same way we conclude that $\left(A, D_{1}, D_{2}\right), \left(B, E_{1}, E_{2}\right)$ and $\left(C, F_{1}, F_{2}\right)$ are collinear.
The corresponding sides of triangles $ABC$ and $D_{i}E_{i}F_{i}$ are perpendicular to each other. Hence, there is a spiral similarity with rotation $\pm 90^{\circ}$ that maps $ABC$ to $D_{i}E_{i}F_{i}$; let $M_{i}$ be the centre of that similarity. Hence, $\Varangle\left(M_{i}A, M_{i}D_{i}\right) = \Varangle\left(M_{i}B, M_{i}E_{i}\right) = \Varangle\left(M_{i}C, M_{i}F_{i}\right) = 90^{\circ}$. The circle with diameter $AD_{i}$ passes through $M_{i}, Y_{i}, Z_{i}$, so $M_{i}, A, Y_{i}, Z_{i}, D_{i}$ are concyclic; analogously ($M_{i}, B, X_{i}, Z_{i}, E_{i}$) and ($M_{i}, C, X_{i}, Y_{i}, F_{i}$) are concyclic.
By applying Desargues' theorem to triangles $ABC$ and $D_{i}E_{i}F_{i}$ we conclude that the lines $AD_{i}, BE_{i}$ and $BF_{i}$ are concurrent; let their intersection be $H$. Since $\left(A, D_{1}, D_{2}\right), \left(B, E_{1}, E_{2}\right)$ and $\left(C, F_{1}, F_{2}\right)$ are collinear, we obtain the same point $H$ for $i=1$ and $i=2$.

By $\Varangle(CB, CH) = \Varangle\left(CX_{i}, CF_{i}\right) = \Varangle\left(Y_{i}X_{i}, Y_{i}F_{i}\right) = \Varangle\left(Y_{i}Z_{i}, Y_{i}D_{i}\right) = \Varangle\left(AZ_{i}, AD_{i}\right) = \Varangle(AB, AH)$, point $H$ lies on circle $ABC$.
Analogously, from $\Varangle\left(F_{i}D_{i}, F_{i}H\right) = \Varangle\left(F_{i}Y_{i}, F_{i}C\right) = \Varangle\left(X_{i}Y_{i}, X_{i}C\right) = \Varangle\left(X_{i}Z_{i}, X_{i}B\right) = \Varangle\left(E_{i}Z_{i}, E_{i}B\right) = \Varangle\left(E_{i}D_{i}, E_{i}H\right)$, we can see that point $H$ lies on circle $D_{i}E_{i}F_{i}$ as well. Therefore, circles $ABC$ and $D_{i}E_{i}F_{i}$ intersect at point $H$.
The spiral similarity moves the circle $ABC$ to circle $D_{i}E_{i}F_{i}$, so the two circles are perpendicular. Hence, both circles $D_{1}E_{1}F_{1}$ and $D_{2}E_{2}F_{2}$ are tangent to the radius of circle $ABC$ at $H$.