Let $$A = \begin{pmatrix} 0 & 2q & r \\ p & q & -r \\ p & -q & r \end{pmatrix}$$. If $$AA^T = I_3$$, then $$|p|$$ is:

We are given the real $$3 \times 3$$ matrix

$$A=\begin{pmatrix}0 & 2q & r\\ p & q & -r\\ p & -q & r\end{pmatrix}$$

and the condition $$AA^T=I_3.$$ For any matrix, the relation $$AA^T=I$$ means that every row of $$A$$ has length $$1$$ (unit vectors) and that any two distinct rows are orthogonal (their dot-product is $$0$$). We therefore write the three rows explicitly:

$$R_1=(0,\;2q,\;r),\qquad R_2=(p,\;q,\;-r),\qquad R_3=(p,\;-q,\;r).$$

First, we impose the unit-length requirement. The square of the length of a vector $$(x_1,x_2,x_3)$$ is $$x_1^{\,2}+x_2^{\,2}+x_3^{\,2}$$. Hence

$$\|R_1\|^2=0^2+(2q)^2+r^2=4q^2+r^2=1,$$

$$\|R_2\|^2=p^2+q^2+(-r)^2=p^2+q^2+r^2=1,$$

$$\|R_3\|^2=p^2+(-q)^2+r^2=p^2+q^2+r^2=1.$$

Next, we apply the orthogonality condition. The dot-product of two vectors $$(x_1,x_2,x_3)$$ and $$(y_1,y_2,y_3)$$ is $$x_1y_1+x_2y_2+x_3y_3$$. Therefore

$$R_1\cdot R_2=0\cdot p+(2q)\,q+r\,(-r)=2q^2-r^2=0,$$

$$R_1\cdot R_3=0\cdot p+(2q)\,(-q)+r\,r=-2q^2+r^2=0,$$

$$R_2\cdot R_3=p\cdot p+q\,(-q)+(-r)\,r=p^2-q^2-r^2=0.$$

We now solve these equations step by step.

From $$2q^2-r^2=0$$ we have

$$r^2=2q^2.$$

Substituting this into the length equation $$4q^2+r^2=1$$ gives

$$4q^2+2q^2=1\quad\Longrightarrow\quad6q^2=1\quad\Longrightarrow\quad q^2=\frac16.$$

Using $$r^2=2q^2$$ again, we find

$$r^2=2\left(\frac16\right)=\frac13.$$

Finally, we substitute $$q^2=\dfrac16$$ and $$r^2=\dfrac13$$ into $$p^2+q^2+r^2=1$$:

$$p^2+\frac16+\frac13=1\quad\Longrightarrow\quad p^2+\frac16+\frac26=1\quad\Longrightarrow\quad p^2+\frac36=1\quad\Longrightarrow\quad p^2+\frac12=1.$$

So

$$p^2=1-\frac12=\frac12,$$

whence

$$|p|=\sqrt{\frac12}=\frac1{\sqrt2}.$$

Hence, the correct answer is Option C.