252. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with edge $a$. Find the length of the shortest segment, the ends of which are located on the lines $A B_{1}$ and $B C_{1}$, forming an angle of $60^{\circ}$ with the plane of the face $A B C D$.
252. Let $M$ be a point on the line $A B_{1}, N$ be on the line $B C_{1}$, $M_{1}$ and $N_{1}$ be the projections of $M$ and $N$ onto the plane $A B C D$. Denote: $\left|B M_{1}\right|=x,\left|B N_{1}\right|=y$, then

$$
\left|M_{1} N_{1}\right|=\sqrt{x^{2}+y^{2}}, \quad|M N|=\sqrt{x^{2}+y^{2}+(a-x-y)^{2}}
$$

By the condition $\left|M N_{r}\right|=2\left|M_{1} N_{1}\right|$, therefore, $(a-x-y)^{2}=$ $=3\left(x^{2}+y^{2}\right)$. Let $x^{2}+y^{2}=u^{2}, x+y=v$, then $2 u^{2}-v^{2} \geqslant 0$, and since $u^{2}=1 / 3(a-v)^{2}$, substituting $u^{2}$ into the inequality relating $u$ and $v$, we get an inequality for $v: v^{2}+4 a c-2 a^{2} \leqslant 0$, from which $-a(2+\sqrt{6}) \leqslant v \leqslant a(\sqrt{6}-2)$. Now let's find the minimum value of $|M N|$, it is equal to $2 a(\sqrt{3}-\sqrt{2})$.