Given two unit vectors $\hat{u}$ and $\hat{v}$, we can construct a vector perpendicular to both by their cross product:
$$\vec{n}=\hat{u}\times\hat{v}.$$
To obtain a perpendicular vector of unit length, just normalize $\vec{n}$:
$$\hat{n}=\frac{\vec{n}}{\|\vec{n}\|}=\frac{\hat{u}\times\hat{v}}{\|\hat{u}\times\hat{v}\|}.$$
Normalizing $\vec{n}$ requires the computation of $\|\hat{u}\times\hat{v}\|$. Since the norm of a vector is defined as the square root of the dot product of the vector with itself, it is impossible to normalize a vector without using square roots.
However, there is a way to look like you're avoiding square roots. If you can find the angle $\theta$ between the unit vectors $\hat{u}$ and $\hat{v}$ geometrically, you can employ the theorem that gives the norm of their cross product as:
$$\|\hat{u}\times\hat{v}\|=\sin{\theta}\\
\implies \hat{n}=\csc{\theta}\,(\hat{u}\times\hat{v}).$$
Of course, this method doesn't truly avoid square roots, since $\sin\theta$ is defined as a square root:
$$|\sin\theta| = \sqrt{1-\cos^2{\theta}}=\sqrt{1-(\hat{u}\cdot\hat{v})^2}.$$
Note that a vector $\vec v$ is orthogonal to your given vectors if and only if $A\vec v=\vec 0$ where
$$
A=
\left[\begin{array}{rrrr}
-1 & 2 & -1 & 0 \\
-3 & 9 & -3 & 2 \\
-1 & -10 & 2 & -8
\end{array}\right]
$$
Thus we wish to compute a basis for the null space of $A$. Row-reducing gives
$$
\DeclareMathOperator{rref}{rref}\rref A=
\left[\begin{array}{rrrr}
1 & 0 & 0 & \frac{4}{3} \\
0 & 1 & 0 & \frac{2}{3} \\
0 & 0 & 1 & 0
\end{array}\right]
$$
Hence a $\vec v$ is orthogonal to your given vectors if and only if
$$
\vec v =
\begin{bmatrix}v_1\\ v_1\\ v_3\\ v_4\end{bmatrix}=
\begin{bmatrix}-\frac{4}{3}\,v_4\\ -\frac{2}{3}\,v_4\\0\\ v_4\end{bmatrix}=
\begin{bmatrix}-\frac{4}{3}\\ -\frac{2}{3}\\0\\ 1\end{bmatrix}v_4
$$
That is, the vectors orthogonal to your given vectors are exactly the scalar multiples of $\left\langle -\frac{4}{3}, -\frac{2}{3}, 0, 1\right\rangle$
Best Answer
$$ n = 3 \cdot \frac{u \times v}{\|u \times v\|}$$