Given $ a_0 \equal{} 1$, $ a_1 \equal{} 3$, and the general relation $ a_n^2 \minus{} a_{n \minus{} 1}a_{n \plus{} 1} \equal{} (\minus{}1)^n$ for $ n \ge 1$. Then $ a_3$ equals:

$ \textbf{(A)}\ \frac{13}{27}\qquad 
\textbf{(B)}\ 33\qquad 
\textbf{(C)}\ 21\qquad 
\textbf{(D)}\ 10\qquad 
\textbf{(E)}\ \minus{}17$
1. We start with the given initial conditions and the recurrence relation:
   \[
   a_0 = 1, \quad a_1 = 3, \quad a_n^2 - a_{n-1}a_{n+1} = (-1)^n \quad \text{for} \quad n \geq 1
   \]

2. For \( n = 1 \):
   \[
   a_1^2 - a_0 a_2 = (-1)^1
   \]
   Substituting the known values \( a_0 = 1 \) and \( a_1 = 3 \):
   \[
   3^2 - 1 \cdot a_2 = -1
   \]
   Simplifying this equation:
   \[
   9 - a_2 = -1
   \]
   Solving for \( a_2 \):
   \[
   a_2 = 10
   \]

3. For \( n = 2 \):
   \[
   a_2^2 - a_1 a_3 = (-1)^2
   \]
   Substituting the known values \( a_1 = 3 \) and \( a_2 = 10 \):
   \[
   10^2 - 3 \cdot a_3 = 1
   \]
   Simplifying this equation:
   \[
   100 - 3a_3 = 1
   \]
   Solving for \( a_3 \):
   \[
   3a_3 = 99
   \]
   \[
   a_3 = 33
   \]

Thus, the value of \( a_3 \) is \( \boxed{33} \).