This method was taught by someone and I am trying to understand the process.
Assume that we have $(1,2,3)$ as first vector. Since we need another vector that will make the dot product $= 0$ with the first vector so let's say the second vector is $(2, -1, 0)$. Now we need a vector orthogonal to $(1,2,3)$ and $(2,-1,0)$ and he said that will be $(b,2b,-\frac53b)$. By making $(1,2,3)$ and $(2,-1,0)$ as unit vectors, then we get $b = 3/70^{1/2}$.
I am curious where $(b,2b,-\frac53b)$ came from.
We have $$ u = (1, 2, 3), \quad v = (2, 1, 0) $$ Testing $w = b (1,2,-5/3)$: $$ u \cdot w = b(1 + 4 - 3*5/3) = 0 \quad v \cdot w = b(2 + 2 + 0) = 4 b $$ So $w$ is orthogonal to $u$ but not $v$, because only the choice $b = 0$ would give an orthogonal vector $w = 0$, but you want a non-zero orthogonal vector. So this was not working.
We can use the vector product to find a vector $w$ orthogonal to both of those vectors: $$ w = u \times v $$ For the $k$-th component this is $$ w_i = \sum_{j,k} \epsilon_{ijk} u_j v_k $$ and gives $$ w = (2\cdot 0 - 3 \cdot 1, 3\cdot 2 - 1\cdot 0, 1\cdot 1 - 2\cdot 2) = (-3, 6, -3) $$
Test: $$ u \cdot w = -3 + 12 - 9 = 0 \quad v \cdot w = -6 + 6 + 0 = 0 $$ Any multiple $b(-3,6,-3)$ for $b \ne 0$ will be orthogonal to $u$ and $v$.