$\therefore 4{q^2} + {r^2} = {p^2} + {q^2} + {r^2} = 1\,\,\,\,\,\,.......\left( 1 \right)$
${p^2} - {q^2} - {r^2} = 0\,\,\,\,\,\,...\left( 2 \right)$
and $2{q^2} - {r^2} = 0\,\,\,\,\,....\left( 3 \right)$
Solving $(1),(2)$ and $(3)$
${p^2} = \frac{1}{2}$
$\left| p \right| = \frac{1}{{\sqrt 2 }}$
$A = \left[ {\begin{array}{*{20}{c}}
{{{10}^{30}} + 5}&{{{10}^{20}} + 4}&{{{10}^{20}} + 6}\\
{{{10}^4} + 2}&{{{10}^8} + 7}&{{{10}^{10}} + 2n}\\
{{{10}^4} + 8}&{{{10}^6} + 4}&{{{10}^{15}} + 9}
\end{array}} \right]$ , $n \in N$, હોય તો . ..