Question - Analysis of a Mathematical Expansion and Counting Problem

The question you've asked appears to be two separate math problems:

1. For the number of words each of 3 vowels and 2 consonants that can be formed from "UTE" - This seems like a permutations problem, but the question seems incomplete as it does not state how many vowels and consonants are available to choose from. Without further information, this part of the question cannot be solved.

2. For finding the 3rd term from the end in the expansion of \(\left(\frac{3}{x^2} - x^3\right)^7\) - This can be solved using the binomial theorem which states that \(n\)th term from the end is given by \(T_{(r+1)} = ^nC_r \cdot a^{(n-r)} \cdot b^r\) for the expansion of \((a + b)^n\), where \(n\) is the power and \(r\) is \(n-k\) if you're looking for the \(k\)th term from the end.

We're looking for the 3rd term from the end (which is the same as the 5th term from the beginning since there are 7 terms in total), so \(r=7-3=4\). Using this, the term is:

\[T_{(4+1)} = ^7C_4 \cdot \left(\frac{3}{x^2}\right)^{(7-4)} \cdot (x^3)^4\]

\[T_5 = ^7C_4 \cdot \frac{3^3}{x^6} \cdot x^{12}\]

\[T_5 = 35 \cdot \frac{27}{x^6} \cdot x^{12}\]

\[T_5 = 35 \cdot 27 \cdot x^6\]

\[T_5 = 945x^6\]

So the 3rd term from the end in the expansion of \(\left(\frac{3}{x^2} - x^3\right)^7\) is \(945x^6\).