1. We are given

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

and $AA^T=I_3.$

This means the rows of $A$ form an orthonormal set.

Let the rows be

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

So we must have:

• $R_i\cdot R_i=1$ for each $i$
• $R_i\cdot R_j=0$ for $i\ne j$

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2. Use orthogonality of different rows.

First,

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

Hence,

$$2q^2-r^2=0 \implies r^2=2q^2.$$

Next,

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

This gives the same condition:

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

Now,

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

Since rows are orthogonal,

$$p^2-q^2-r^2=0.$$

Using $r^2=2q^2$,

$$p^2-q^2-2q^2=0 \implies p^2=3q^2.$$

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3. Use normalization of one row.

From $R_1\cdot R_1=1$,

$$0^2+(2q)^2+r^2=1 \implies 4q^2+r^2=1.$$

Using $r^2=2q^2$,

$$4q^2+2q^2=1 \implies 6q^2=1 \implies q^2=\frac16.$$

Then

$$p^2=3q^2=3\cdot \frac16=\frac12.$$

Therefore,

$$|p|=\sqrt{\frac12}=\frac{1}{\sqrt2}.$$

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4. Check with options.

$|p|=\frac{1}{\sqrt2}$ which is option A.

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5. Comparison with stored correct answer.

Stored correct answer: A

Our derived answer also gives A, so they agree.