Let the first row have $n$ children and total rows be $r$. Then the numbers per row form an AP with difference $-3$ and sum $630$:
$\displaystyle \frac{r}{2}\big(2n-3(r-1)\big)=630$ $ \;\Rightarrow\; 1260=r\big(2n-3(r-1)\big)$.
Hence $r\mid1260$. Also the last row must be positive: $n-3(r-1)>0$.
Testing the options (and ensuring $n$ is integer and last term positive):
Therefore, the impossible number of rows is $\boxed{6}$.
Arrangement:
1 A, 2 N, 3 B, 4 O, 5 C, 6 P, 7 D, 8 Q, 9 E, 10 R, 11 F, 12 S, 13 G, 14 T, 15 H, 16 U, 17 I, 18 V, 19 J, 20 W, 21 K, 22 X, 23 L, 24 Y, 25 M, 26 Z
Step:
7th letter from the left = D (position 7)
3rd letter to the right of D = position 10 = R
Answer: R
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and More.