105. All plane angles of the trihedral angle $N K L M (N-$ vertex) are right angles. On the face $L N M$, a point $P$ is taken at a distance of 2 from the vertex $N$ and at a distance of 1 from the edge $M N$. A light ray is directed from some point $S$, located inside the trihedral angle $N K L M$, to the point $P$. The ray forms an angle $\pi / 4$ with the plane $M N K$ and equal angles with the edges $K N$ and $M N$. The ray is mirror-reflected from the faces of the angle $N K L M$ first at point $P$, then at point $Q$, and then at point $R$. Find the sum of the lengths of segments $P Q$ and $Q R$.
105. The solution of the problem is based on the fact that the continuation of the falling ray is symmetric to the reflected ray relative to the edge from which the ray is reflected. We will introduce a coordinate system naturally, taking its origin at point $N$, and as the $x, y$ and $z$ axes, the edges $N K, N L$ and $N M$, and denote by $Q''$ and $R''$ the subsequent points of intersection of the line $S P$ with the coordinate planes, different from $L N M$. We have: $|P Q|=\left|P Q^{\prime}\right|,|Q R|=\left|Q^{\prime} R^{\prime \prime}\right|$.

Point $P$ has coordinates $(0,1, \sqrt{3})$. Let $\alpha, \beta, \alpha$ be the angles formed by the ray $S P$ with the coordinate axes. From the condition, it follows that $\beta=\pi / 4$, and then we find $\cos \alpha$ from the equality $2 \cos ^{2} \alpha+$ $+\cos ^{2} \beta=1, \cos \alpha=1 / 2$ ( $\alpha$ - acute angle). Therefore, the vector $\boldsymbol{a}(1 / 2, \sqrt{2} / 2,1 / 2)$ is parallel to the line $S P$. If $A(x, y, z)$ is an arbitrary point on this line, then

$$
\overrightarrow{O A}=\overrightarrow{O P}+t a
$$

or coordinate-wise

$$
x=\frac{t}{2}, \quad y=1+\frac{\sqrt{2}}{2} t, \quad z=\sqrt{3}+\frac{t}{2}
$$

The coordinates $y$ and $z$ become zero at $t_{1}=-\sqrt{2}$ (this will be point $Q^{\prime}$) and at $t_{2}=-2 \sqrt{3}$ (point $R^{\prime}$). Thus, $Q^{\prime}\left(-\frac{\sqrt{\overline{2}}}{2}\right.$,

$$
\begin{aligned}
& \left.0, \sqrt{3}-\frac{\sqrt{\overline{2}}}{2}\right), R^{\prime}(-\sqrt{\overline{3}}, 1-\sqrt{\overline{6}}, 0) \\
& \left|P Q^{n}\right|=\sqrt{2}, \quad\left|Q^{\prime} R^{\prime}\right|=2 \sqrt{\overline{3}}-\sqrt{2}
\end{aligned}
$$

Answer: $2 \sqrt{3}$.