Let A= ( … ). If AA^ T= I_3, then | | is:

$A=\left[\begin{array}{ccc}0 & 2 q & r \\ p & q & -r \\ p & -q & r\end{array}\right]$A=​0pp​2qq−q​r−rr​​ $\begin{aligned} \therefore & A \cdot A^{T}=\left[\begin{array}{ccc}0 & 2 q & r \\ p & q & -r \\ p & -q & r\end{array}\right] \times\left[\begin{array}{ccc}0 & p & p \\ 2 q & q & -q \\ r & -r & r\end{array}\right] \\ &=\left[\begin{array}{ccc}4 q^{2}+r^{2} & 2 q^{2}-r^{2} & -2 q^{2}+r^{2} \\ 2 q^{2}-r^{2} & p^{2}+q^{2}+r^{2} & p^{2}-q^{2}-r^{2} \\ -2 q^{2}+r^{2} & p^{2}-q^{2}-r^{2} & p^{2}+q^{2}+r^{2}\end{array}\right] \end{aligned}$∴​A⋅AT=​0pp​2qq−q​r−rr​​×​02qr​pq−r​p−qr​​=​4q2+r22q2−r2−2q2+r2​2q2−r2p2+q2+r2p2−q2−r2​−2q2+r2p2−q2−r2p2+q2+r2​​​ Given, $A A^{T}=I$AAT=I $\therefore \quad 4 q^{2}+r^{2}=p^{2}+q^{2}+r^{2}=1$∴4q2+r2=p2+q2+r2=1 $\Rightarrow \quad p^{2}-3 q^{2}=0$⇒p2−3q2=0 and $r^{2}=1-4 q^{2}$r2=1−4q2 and $2 q^{2}-r^{2}=0 \Rightarrow r^{2}=2 q^{2}$2q2−r2=0⇒r2=2q2 $\begin{array}{ll}\therefore \quad & p^{2}=\frac{1}{2}, q^{2}=\frac{1}{6} \text { and } r^{2}=\frac{1}{3} \\ \therefore \quad & |p|=\frac{1}{\sqrt{2}}\end{array}$∴∴​p2=21​,q2=61​ and r2=31​∣p∣=2​1​​