Answer: a) $\sqrt{2017^{2}+2}$, b) it exists. Solution. a) Let the dimensions of the parallelepiped be $a \times b \times c$. Then we have $a b c=2017$, and due to the simplicity of the number 2017, we get that the edges $a, b, c$ are (in some order) $2017,1,1$, so the diagonal is $\sqrt{2017^{2}+1^{2}+1^{2}}$. b) 2017 is a prime number of the form $4 \mathrm{k}+1$, and such numbers can be represented as the sum of two perfect squares; in this case $2017=1936+81=44^{2}+9^{2}$ (such a representation is not difficult to find, since 44 is the largest natural number whose square is less than 2017). Therefore, we can construct a parallelepiped whose base is a square in the xy-plane with sides not parallel to the x and y axes, and whose height is the segment $[0 ; 1]$ of the z-axis (the vertices of the square are the points $(0 ; 0),(44 ; 9),(-9 ; 44)$, $(35 ; 53)$, the side of the square is $\sqrt{2017})$. The diagonal of this parallelepiped is $\sqrt{2017+2017+1}=$ $\sqrt{4035}$
11.4. a) Given a rectangular parallelepiped with a volume of 2017 and integer coordinates of vertices in a Cartesian coordinate system in space. Find the diagonal of the parallelepiped, given that its edges are parallel to the coordinate axes. b) Does there exist a rectangular parallelepiped with a volume of 2017 and integer coordinates of vertices, where not all edges are parallel to the coordinate axes.