1.23. Inside a sphere of radius $R$, a point $A$ is taken at a distance $a$ from its center. Through point $A$, three pairwise perpendicular chords are drawn.

a) Find the sum of the squares of the lengths of these chords.

b) Find the sum of the squares of the lengths of the segments into which point $A$ divides these chords.
1.23. Consider a rectangular parallelepiped, the edges of which are parallel to the given chords, and points $A$ and the center $O$ of the sphere are its opposite vertices. Let $a_{1}, a_{2}$, and $a_{3}$ be the lengths of its edges; it is clear that $a_{1}^{2}+a_{2}^{2}+a_{3}^{2}=a^{2}$.

a) If a chord is at a distance $x$ from the center of the sphere, then the square of its length is $4 R^{2}-4 x^{2}$. Since the distances from the given chords to point $O$ are the diagonals of the faces of the parallelepiped, the desired sum of squares is $12 R^{2}-4\left(a_{2}^{2}+a_{3}^{2}\right)$ ? $-4\left(a_{1}^{2}+a_{3}^{2}\right)-4\left(a_{1}^{2}+a_{2}^{2}\right)=12 R^{2}-8 a^{2}$

b) If the length of the chord is $d$, and point $A$ is at a distance $y$ from its midpoint, then the sum of the squares of the lengths of the segments into which the chord is divided by point $A$ is $2 y^{2}+d^{2} / 2$. Since the distances from point $A$ to the midpoints of the given chords are $a_{1}, a_{2}$, and $a_{3}$, and the sum of the squares of the lengths of the chords is $12 R^{2}-8 a^{2}$, the desired sum of squares is $2 a^{2}+\left(6 R^{2}-4 a^{2}\right)=6 R^{2}-2 a^{2}$.