12*. Given perpendicular intersecting lines $a$ and $b$ and a point $P$. What does the set of points $M$ represent, for which the sum of the lengths of the projections of the segment $P M$ onto the lines $a$ and $b$ is constant? [The projection of a point $A$ onto a line $l$ is a point $A^{\prime} \in l$ such that $\left(A A^{\prime}\right) \perp l$; if $A \in l$, then $A^{\prime}=A$. The projection of a segment $A B$ onto a line $l$ is a segment $A^{\prime} B^{\prime}$, where points $A^{\prime}$ and $B^{\prime}$ are the projections of points $A$ and $B$ onto the line $l$.]
12. Draw through $P$ lines $a^{\prime} \| a$ and $b^{\prime} \| b$. These lines lie in a plane $\alpha$, parallel to both lines. Let the projection of segment $P M$ onto plane $\alpha$ be segment $P M^{\prime}$ (Fig. 85). Since the projections of segment $P M$ onto lines $a$ and $b$ are parallel to plane $\alpha$, they are congruent to their projections onto this plane, and also congruent to the projections of segment $P M^{\prime}$ onto lines $a^{\prime}$ and $b^{\prime}$. Our problem is reduced to a planimetric one: what is the set of points $M^{\prime}$ in plane $\alpha$, the sum of whose distances to two mutually perpendicular lines $a^{\prime}$ and $b^{\prime}$ is equal to a given segment $m$?

To solve this problem, we take $a^{\prime}$ and $b^{\prime}$ as the coordinate axes with the origin at point $P$. Then the position of point $M^{\prime}$ is determined by the equation: $|x| + |y| = m$.

This equation defines the union of four segments in the four coordinate quadrants, forming a square (see Fig. 85). Therefore, point $M$ lies on the perpendicular to plane $\alpha$ drawn through one of the points of these segments, and the desired set of points is the union of 4 strips, the planes of which are perpendicular to $\alpha$.