To determine the correct condition when there is a term containing \(x^{2r}\) in the expansion of \(\left(x + \frac{1}{x^2}\right)^{n-3}\), we will analyze the binomial expansion step-by-step.
The expression \(\left(x + \frac{1}{x^2}\right)^{n-3}\) can be expanded using the Binomial Theorem:
\[\left(x + \frac{1}{x^2}\right)^{n-3} = \sum_{k=0}^{n-3} \binom{n-3}{k} x^{(n-3)-k} \left(\frac{1}{x^2}\right)^k\]Simplifying the expression inside the summation:
\[= \sum_{k=0}^{n-3} \binom{n-3}{k} x^{(n-3-k) - 2k} = \sum_{k=0}^{n-3} \binom{n-3}{k} x^{n-3 - 3k}\]For a term with \(x^{2r}\), set the power of \(x\) equal to \(2r\):
\[n-3 - 3k = 2r\]Rearranging the equation, we get:
\[n - 2r = 3k + 3\]This implies that \(n - 2r\) must be a positive integral multiple of 3 as it can be expressed as \(3(k + 1)\), where \(k \geq 0\).
Therefore, the correct answer is that \(n - 2r\) is a positive integral multiple of 3.