7. In space, three distinct points $X, Y$ and $Z$ are given; it is asked whether there exists a point $P$ different from $X, Y$ and $Z$ such that the lines $P X, P Y$ and $P Z$ are pairwise perpendicular. Four friends make the following statements:

Alberto: "There exist $X, Y, Z$ and $P$ belonging to the same plane that satisfy these conditions." Barbara: "There do not exist 4 points $X, Y, Z$ and $P$ that satisfy these conditions." Carlo: "For $P$ to exist, it is necessary that $X Y Z$ be acute-angled." Daria: "For $P$ to exist, it is sufficient that $X Y Z$ not be obtuse-angled."

Who is right?
(A) Only Alberto
(B) Only Barbara
(C) Only Carlo
(D) Only Daria
(E) None of the above.
7. The answer is $(\mathbf{C})$. The only one who is right is Carlo.

Alberto is wrong, indeed if there existed $X, Y, Z, P$ belonging to the same plane such that $P X, P Y$ and $P Z$ are pairwise perpendicular, then $P Z \perp P X, P Z \perp P Y$ and thus $P Y \| P X$, which is absurd.

Barbara is wrong: 4 such points are for example given by the endpoints of 3 edges of a cube having a vertex in common (which will be $P$), or more generally by the vertices of any right-angled tetrahedron.

Carlo is right: indeed, since $X Y^{2}=P X^{2}+P Y^{2}<P X^{2}+P Y^{2}+2 \cdot P Z^{2}=X Z^{2}+Y Z^{2}$, then $\widehat{X Z Y}$ is acute, and similarly $\widehat{Z X Y}$ and $\widehat{X Y Z}$ are also acute. An alternative argument to show that Carlo is right is as follows: consider the sphere with diameter $X Y$, then $P$ belongs to this sphere and the line $P Z$ is perpendicular to the plane on which $P, X, Y$ lie, hence any point on this line (except for $P$) is outside the sphere. Therefore, $\widehat{X Z Y}$ is acute (and similarly $\widehat{Z X Y}$ and $\widehat{X Y Z}$ are acute).

Daria is wrong: as stated above, it is necessary that $X Y Z$ be acute-angled, so $X Y Z$ cannot be a right triangle; in particular, it is not sufficient that $X Y Z$ is not obtuse.