$L e t A = [\begin{matrix} a & x & p \\ y & q & b \\ r & c & z \end{matrix}] a n d B = [\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}]$ where a,b,cx,y,z,p,q,r are natural numbers.
If tr. $(A B + A B^{3} + A B^{5} + .... + A B^{19}) = 210$ , Then if N = the number of ordered triplets $(p, q, r)$
Then number of divisors of N is
[Note:tr.(P) denotes the trace of matrix P.]

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Detailed Solution

$B^{2} = I$
$A B = [\begin{matrix} a & x & p \\ y & q & b \\ r & c & z \end{matrix}] [\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}] = [\begin{matrix} p & x & a \\ b & q & y \\ z & c & r \end{matrix}]$
$A B = A B^{3} = ...... = A B^{19} = [\begin{matrix} p & x & a \\ b & q & y \\ z & c & r \end{matrix}]$
$t r . (A B + A B^{3} + ...... + A B^{19}) = 210$
$\Rightarrow 10 (p + q + r) = 210 \Rightarrow p + q + r = 21, p, q, r \in N$
$p' + q' + r' = 18, p', q', r' \in W$
$∴$ Number of ordered triplets $(p, q, r) =^{20} C_{2} = \frac{20 \times 19}{2} = 190$