Это решается через определитель:
V = \frac{1}{6} \begin{vmatrix}
X(B)-X(A) & Y(B)-Y(A) & Z(B)-Z(A) \\
X(C)-X(A) & Y(C)-Y(A) & Z(C)-Z(A) \\
X(D)-X(A) & Y(D)-Y(A) & Z(D)-Z(A) \\
\end{vmatrix} =
= \frac{1}{6} \begin{vmatrix}
2-1 & 0-1 & 2-1 \\
2-1 & 2-1 & 2-1 \\
3-1 & 4-1 & -3-1 \\
\end{vmatrix} = \frac{1}{6} \begin{vmatrix}
1 & -1 & 1 \\
1 & 1 & 1 \\
2 & 3 & -4 \\
\end{vmatrix} =
= 1/6·|1·1(–4) + 1·1·3 + 2·1(–1) – 1·1·2 – 1·1·3 – 1(–1)(–4)| =
= 1/6·|–4 + 3 – 2 – 2 – 3 – 4| = 1/6·|–12| = 1/6·12 = 2
Ответ: 2