Solution:

Let $\omega_1$ denote the circumcircle of $ABC$ and $\omega_2$ denote the circle centered at $X$ through $B$ and $C$. Let $\omega_2$ intersect $AB$, $AC$ again at $B'$, $C'$. The (signed) power of $Y$ with respect to $\omega_1$ is $-CY \cdot YZ$. The power of $Y$ with respect to $\omega_2$ is $XY^2 - CX^2 = -CY^2$. Thus the ratio of the powers of $Y$ with respect to the two circles is $3 : 1$. The circumcircle of $BCY$ passes through the intersection points of $\omega_1$ and $\omega_2$ ($B$ and $C$) and thus contains exactly the set of points such that the ratio of their powers with respect to $\omega_1$ and $\omega_2$ is $3 : 1$ (this fact can be verified in a variety of ways). We conclude that $K$ must be the point on line $AB$ such that $\frac{KB'}{KA} = 3$. It now suffices to compute $AB'$. Note $AB' = \frac{AC \cdot B'C'}{BC}$ by similar triangles. Also an angle chase gives that $B'XC'$ are collinear. We compute

$$BX = \frac{BC}{2 \cos A} = \frac{5}{2 \cdot \frac{5}{7}} = \frac{7}{2}$$
and thus $B'C' = 7$ so $AB' = \frac{49}{5}$ and $AK = \frac{3}{2} AB' = \frac{147}{10}$.