Is there are a $3 \times 3$ real, orthogonal matrix $Q$ that has exactly three zero entries?

Question

If I have choose first column with two zeros, then other two columns have two or none.

Or else there are are three zeros in each column. I don't know how to conclude after this

Answer 1

Indeed, every $3\times 3$ matrix where the first column has two zeros cannot be orthogonal. Why? Orthogonality means that every two column vectors are perpendicular to each other, i.e. their scalar product is $0$. But if one column vector $u$ just has one non-zero value (wlog it's the first entry), we have the following equation: $$\langle u,v\rangle=u_1\cdot v_1 + 0\cdot v_2 + 0\cdot v_3 \overset{!}{=} 0$$ with every other column vector $v$. That means that the other two column vectors needs to have a $0$ as the first entry resulting in a total of $4$ zero entries.

For an analogous reason it is not possible to let each column vector have one zero-entry: wlog let's $u_1=0,v_2=0$, then $$ \langle u,v\rangle=0\cdot v_1 + u_2\cdot 0+ u_3\cdot v_3 \overset{!}{=} 0 $$ so $u_3=0$ or $v_3=0$ and we have the first case again.

If you have $3$ zeros in one column or row your matrix is not invertible (so also not orthogonal) anymore.

Concluded, it is not possible to get a $3\times 3$ matrix with exactly $3$ zero entries!