This was somewhat long so I wouldn't be surprised if there's an error in here somewhere so read critically.

Let $A = \begin{bmatrix}

a_1 & a_2 & a_3 \\

b_1 & b_2 & b_3 \\

c_1 & c_2 & c_3

\end{bmatrix}$

and $v= \begin{bmatrix}

v_1 \\

v_2\\

v_3

\end{bmatrix}$

Two vectors are othogonal if their dot product is $0$, so we know $Av \cdot v = 0$

$Av = \begin{bmatrix}

a_1 v_1 + a_2 v_2 + a_3 v_3 \\

b_1 v_1 + b_2 v_2 + b_3 v_3 \\

c_1 v_1 + c_2 v_2 + c_3 v_3

\end{bmatrix}$

$Av \cdot v = v_1 (a_1 v_1 + a_2 v_2 + a_3 v_3)\\

+ v_2 (b_1 v_1 + b_2 v_2 + b_3 v_3)\\

+ v_3 (c_1 v_1 + c_2 v_2 + c_3 v_3) = 0$

after a bit of simplifying:

$Av \cdot v = a_1 v_{1}^{2} + b_2 v_{2}^{2} + c_3 v_{3}^{2} + v_1 v_2 (a_2 + b_1) + v_1 v_3 (a_3 + c_1) + v_2 v_3 (b_3 + c_2) = 0$

So , if $a_1 , b_2 , c_3 = 0$ and $b_1 = -a_2 $ , $c_1 = -a_3$ , and $c_2 = -b_3$ then $Av \cdot v = 0$

So $A=\begin{bmatrix}

0 & a_2 & a_3 \\

-a_2 & 0 & b_3 \\

-a_3 & -b_3 & 0

\end{bmatrix}$

And you then have $A^{t} + A = 0$