Let's first prove the following statement: if the edges of a triangular pyramid are pairwise perpendicular, then the square of the area of the base is equal to the sum of the squares of the areas of the lateral faces. Indeed, let \( OX, OY \) and \( OZ \) be the pairwise perpendicular lateral edges of the triangular pyramid \( OXYZ \) with vertex \( O \) (Fig.1), and let \( S_{\Delta} X O Y = S, S_{\triangle X O Z} = P, S_{\triangle Y O Z} = Q \). Denote \( OX = a, OY = b, OZ = c \). Then

\[
\left\{\begin{array}{l}
S = \frac{1}{2} a b \\
P = \frac{1}{2} a c \\
Q = \frac{1}{2} b c
\end{array}\right.
\]

Multiplying the first two equations term by term and dividing the result by the third, we get \( a = \sqrt{Q} \). Let \( S_{\triangle X Y Z} = T \). We will prove that \( T^2 = S^2 + P^2 + Q^2 \). For this, drop the perpendicular \( OF \) from vertex \( O \) to edge \( YZ \). Edge \( OX \) is perpendicular to the plane of face \( OYZ \) because \( OX \perp OY \) and \( OX \perp OZ \) by the problem's condition. Then line \( YZ \) is perpendicular to the plane \( OXF \), meaning \( XF \perp YZ \), i.e., \( XF \) is the height of triangle \( XYZ \). From the right triangles \( YOZ \) and \( XOF \), we find that

\[
O F = \frac{O Y \cdot O Z}{Y Z} = \frac{O Y \cdot O Z}{\sqrt{O Y^2 + O Z^2}} = \frac{b c}{\sqrt{b^2 + c^2}}
\]

\[
X F = \sqrt{O X^2 + O F^2} = \sqrt{a^2 + \frac{b^2 c^2}{b^2 + c^2}} = \frac{\sqrt{a^2 b^2 + a^2 c^2 + b^2 c^2}}{\sqrt{b^2 + c^2}}
\]

Thus,

\[
\begin{gathered}
T = S_{\triangle X Y Z} = \frac{1}{2} Y Z \cdot X F = \frac{1}{2} \sqrt{b^2 + c^2} \cdot \frac{\sqrt{a^2 b^2 + a^2 c^2 + b^2 c^2}}{\sqrt{b^2 + c^2}} = \\
= \frac{1}{2} \sqrt{a^2 b^2 + a^2 c^2 + b^2 c^2} = \sqrt{\frac{1}{4} a^2 b^2 + \frac{1}{4} a^2 c^2 + \frac{1}{4} b^2 c^2} = \sqrt{S^2 + P^2 + Q^2}
\end{gathered}
\]

Therefore, \( T^2 = S^2 + P^2 + Q^2 \). This is what we needed to prove. Now let's move on to our problem (Fig.2). Draw a section of the sphere with the plane \( DA_1 B_1 \). We get a circle and two secants \( DA_1 A_2 \) and \( DB_1 B_2 \) drawn to it from point \( D \), which lies outside the circle. Then \( DA_1 \cdot DA_2 = DB_1 \cdot DB_2 \). Similarly, we can prove that \( DA_1 \cdot DA_2 = DC_1 \cdot DC_2 \). Moreover,

\[
\begin{aligned}
& \frac{S_{\Delta D A_1 B_1}}{S_{\Delta D B_1 C_1}} = \frac{D A_1 \cdot D B_1}{D B_1 \cdot D C_1} = \frac{D A_1}{D C_1}, \\
& \frac{S_{\Delta D B_2 C_2}}{S_{\Delta D A_2 B_2}} = \frac{D B_2 \cdot D C_2}{D A_2 \cdot D B_2} = \frac{D C_2}{D A_2}
\end{aligned}
\]

Thus,

\[
\frac{S_{\Delta D A_1 B_1}}{S_{\Delta D B_1 C_1}} = \frac{S_{\Delta D B_2 C_2}}{S_{\Delta D A_2 B_2}},
\]

from which

\[
S_{\Delta D B_2 C_2} = S_{\Delta D A_2 B_2} \cdot \frac{S_{\Delta D A_1 B_1}}{S_{\Delta D B_1 C_1}} = 40 \cdot \frac{\frac{1}{6}}{6} = 50
\]

Similarly,

\[
S_{\Delta D A_2 C_2} = S_{\Delta D A_2 B_2} \cdot \frac{S_{\Delta D A_1 B_1}}{S_{\Delta D A_1 C_1}} = 40 \cdot \frac{\frac{1}{10}}{\frac{2}{10}} = 30
\]

Therefore,

\[
\begin{gathered}
S_{\Delta A_2 B_2 C_2} = \sqrt{S_{\Delta D A_2 B_2}^2 + S_{\Delta D B_2 C_2}^2 + S_{\Delta D A_2 C_2}^2} = \sqrt{40^2 + 50^2 + 30^2} = \\
= 10 \sqrt{16 + 25 + 9} = 10 \sqrt{50} = 50 \sqrt{2}
\end{gathered}
\]

Answer

\( 50 \sqrt{2} \).
In space, three rays $D A, D B$, and $D C$ are given, having a common origin $D$, and $\angle A D B=\angle A D C=\angle B D C=$ $90^{\circ}$. A sphere intersects the ray $D A$ at points $A 1$ and $A 2$, the ray $D B$ - at points $B 1$ and $B 2$, and the ray $D C$ - at points $C 1$ and $C 2$. Find the area of triangle $A 2 B 2 C 2$, if the areas of triangles $D A 1 B 1, D A 1 C 1, D B 1 C 1$ and $D A 2 B 2$ are respectively $\frac{\frac{15}{2}}{2}, 10,6$ and 40.