154. In a regular tetrahedron \(ABCD\) with edge \(a\), points \(A_1, B_1, C_1,\) and \(D_1\) are taken in the planes \(BCD, CDA, DAB,\) and \(ABC\) respectively, such that the line \(A_1 B_1\) is perpendicular to the plane \(BCD\), \(B_1 C_1\) is perpendicular to the plane \(CDA\), \(C_1 D_1\) is perpendicular to the plane \(DAB\), and finally, \(D_1 A_1\) is perpendicular to the plane \(ABC\). Find the volume of the tetrahedron \(A_1 B_1 C_1 D_1\).
154. Drop perpendiculars $A_{1} M$ and $B_{1} M$ to $C D$, $B_{1} N$ and $C_{1} N$ to $A D$, $C_{1} K$ and $D_{1} K$ to $A B$, $D_{1} L$ and $A_{1} L$ to $C B$.

Since

$$
\frac{\left|A_{1} M\right|}{\left|B_{1} M\right|}=\frac{\left|B_{1} N\right|}{\left|N C_{1}\right|}=\frac{\left|C_{1} K\right|}{\left|K D_{1}\right|}=\frac{\left|D_{1} L\right|}{\left|A_{1} L\right|}=\frac{1}{3}
$$

(these ratios are equal to the cosine of the dihedral angle at the edge of the tetrahedron) and $\left|A_{1} B_{1}\right|=\left|B_{1} C_{1}\right|=\left|C_{1} D_{1}\right|=\left|D_{1} A_{1}\right|$, the equalities $\left|A_{1} M\right|=\left|B_{1} N\right|=\left|C_{1} K\right|=\left|D_{1} L\right|=$ $=x,\left|B_{1} M\right|=\left|N C_{1}\right|=\left|K D_{1}\right|=\left|A_{1} L\right|=3 x$ (the net of the tetrahedron is shown in Fig. 33) must hold. Each edge $C D, D A, A B, B C$ will be divided into segments $m$ and $n$ as shown in the figure. Considering that $m+n=a$, we find $x=\frac{a \sqrt{3}}{12}, m=\frac{5 a}{12}, n=\frac{7 a}{12}$, after which we will find the volume of the tetrahedron $A_{1} B_{1} C_{1} D_{i}$.

$$
\text { Answer: } \frac{a^{3} \sqrt{2}}{162} .
$$