Consider the polynomial ring $R=\mathbb C[x_1,x_2,...,x_{16}]$, and set
$$X=\begin{pmatrix} x_1 &x_2&x_3 &x_4\\ x_5&x_6& x_7&x_8\\x_9&x_{10}&x_{11}&x_{12}\\x_{13}&x_{14}&x_{15}&x_{16}\end{pmatrix}.$$
Now, using these three matrices
$$L=\begin{pmatrix}0&-1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&1&0 \end{pmatrix}$$ $$M=\begin{pmatrix}0&0&0&-1\\0&0&-1&0\\0&1&0&0\\1&0&0&0\end{pmatrix}$$ $$N=\begin{pmatrix}0&0&-1&0\\0&0&0&1\\1&0&0&0\\0&-1&0&0\end{pmatrix}$$
we create polynomials $f_i, g_i,$ and $h_i$ in the following way:
$$XLX^t-L=\begin{pmatrix} f_1 &f_2&f_3 &f_4\\ f_5&f_6& f_7&f_8\\f_9&f_{10}&f_{11}&f_{12}\\f_{13}&f_{14}&f_{15}&f_{16}\end{pmatrix}$$
$$XMX^t-M=\begin{pmatrix} g_1 &g_2&g_3 &g_4\\ g_5&g_6& g_7&g_8\\g_9&g_{10}&g_{11}&g_{12}\\g_{13}&g_{14}&g_{15}&g_{16}\end{pmatrix}$$
$$XNX^t-N=\begin{pmatrix} h_1 &h_2&h_3 &h_4\\ h_5&h_6& h_7&h_8\\h_9&h_{10}&h_{11}&h_{12}\\h_{13}&h_{14}&h_{15}&h_{16}\end{pmatrix}$$
Finally, let $I = (f_i, g_i, h_i)$ be the ideal generated by these $48$ polynomials. Then how to show that the radical of $I$, i.e. $\sqrt I$, is generated by twelve linear polynomials and one quadratic polynomial ?
I have no idea how to approach this problem; may be use Nullstelensatz ... ?
Please help
NOTE : All the matrices $L,M,N$ are orthogonal , so the three defining equations can be written as $(XL)(LX)^t=(XM)(MX)^t=(XN)(NX)^t=Id$. Now if we can find some pattern in $XL,LX,MX,XM,NX,XN$ then it could be helpful to find the zero set of the ideal $I$ ... Also $L,M,N$ are skew symmetric matrices and as @Balaji sb noted, $LM=-N$ ... this means $L,M,N$ works as the $i,j,k$ in the Quaternion ring ...
Too long for a comment. In M2:
$0,{x}_{2} {x}_{5}-{x}_{1} {x}_{6}+{x}_{4} {x}_{7}-{x}_{3} {x}_{8}+1,{x}_{2} {x}_{9}-{x}_{1} {x}_{10}+{x}_{4} {x}_{11}-{x}_{3} {x}_{12},{x}_{2} {x}_{13}-{x}_{1} {x}_{14}+{x}_{4} {x}_{15}-{x}_{3} {x}_{16},-{x}_{2} {x}_{5}+{x}_{1} {x}_{6}-{x}_{4} {x}_{7}+{x}_{3} {x}_{8}-1,0,{x}_{6} {x}_{9}-{x}_{5} {x}_{10}+{x}_{8} {x}_{11}-{x}_{7} {x}_{12},{x}_{6} {x}_{13}-{x}_{5} {x}_{14}+{x}_{8} {x}_{15}-{x}_{7} {x}_{16},-{x}_{2} {x}_{9}+{x}_{1} {x}_{10}-{x}_{4} {x}_{11}+{x}_{3} {x}_{12},-{x}_{6} {x}_{9}+{x}_{5} {x}_{10}-{x}_{8} {x}_{11}+{x}_{7} {x}_{12},0,{x}_{10} {x}_{13}-{x}_{9} {x}_{14}+{x}_{12} {x}_{15}-{x}_{11} {x}_{16}+1,-{x}_{2} {x}_{13}+{x}_{1} {x}_{14}-{x}_{4} {x}_{15}+{x}_{3} {x}_{16},-{x}_{6} {x}_{13}+{x}_{5} {x}_{14}-{x}_{8} {x}_{15}+{x}_{7} {x}_{16},-{x}_{10} {x}_{13}+{x}_{9} {x}_{14}-{x}_{12} {x}_{15}+{x}_{11} {x}_{16}-1,0,0,{x}_{4} {x}_{5}+{x}_{3} {x}_{6}-{x}_{2} {x}_{7}-{x}_{1} {x}_{8},{x}_{4} {x}_{9}+{x}_{3} {x}_{10}-{x}_{2} {x}_{11}-{x}_{1} {x}_{12},{x}_{4} {x}_{13}+{x}_{3} {x}_{14}-{x}_{2} {x}_{15}-{x}_{1} {x}_{16}+1,-{x}_{4} {x}_{5}-{x}_{3} {x}_{6}+{x}_{2} {x}_{7}+{x}_{1} {x}_{8},0,{x}_{8} {x}_{9}+{x}_{7} {x}_{10}-{x}_{6} {x}_{11}-{x}_{5} {x}_{12}+1,{x}_{8} {x}_{13}+{x}_{7} {x}_{14}-{x}_{6} {x}_{15}-{x}_{5} {x}_{16},-{x}_{4} {x}_{9}-{x}_{3} {x}_{10}+{x}_{2} {x}_{11}+{x}_{1} {x}_{12},-{x}_{8} {x}_{9}-{x}_{7} {x}_{10}+{x}_{6} {x}_{11}+{x}_{5} {x}_{12}-1,0,{x}_{12} {x}_{13}+{x}_{11} {x}_{14}-{x}_{10} {x}_{15}-{x}_{9} {x}_{16},-{x}_{4} {x}_{13}-{x}_{3} {x}_{14}+{x}_{2} {x}_{15}+{x}_{1} {x}_{16}-1,-{x}_{8} {x}_{13}-{x}_{7} {x}_{14}+{x}_{6} {x}_{15}+{x}_{5} {x}_{16},-{x}_{12} {x}_{13}-{x}_{11} {x}_{14}+{x}_{10} {x}_{15}+{x}_{9} {x}_{16},0,0,{x}_{3} {x}_{5}-{x}_{4} {x}_{6}-{x}_{1} {x}_{7}+{x}_{2} {x}_{8},{x}_{3} {x}_{9}-{x}_{4} {x}_{10}-{x}_{1} {x}_{11}+{x}_{2} {x}_{12}+1,{x}_{3} {x}_{13}-{x}_{4} {x}_{14}-{x}_{1} {x}_{15}+{x}_{2} {x}_{16},-{x}_{3} {x}_{5}+{x}_{4} {x}_{6}+{x}_{1} {x}_{7}-{x}_{2} {x}_{8},0,{x}_{7} {x}_{9}-{x}_{8} {x}_{10}-{x}_{5} {x}_{11}+{x}_{6} {x}_{12},{x}_{7} {x}_{13}-{x}_{8} {x}_{14}-{x}_{5} {x}_{15}+{x}_{6} {x}_{16}-1,-{x}_{3} {x}_{9}+{x}_{4} {x}_{10}+{x}_{1} {x}_{11}-{x}_{2} {x}_{12}-1,-{x}_{7} {x}_{9}+{x}_{8} {x}_{10}+{x}_{5} {x}_{11}-{x}_{6} {x}_{12},0,{x}_{11} {x}_{13}-{x}_{12} {x}_{14}-{x}_{9} {x}_{15}+{x}_{10} {x}_{16},-{x}_{3} {x}_{13}+{x}_{4} {x}_{14}+{x}_{1} {x}_{15}-{x}_{2} {x}_{16},-{x}_{7} {x}_{13}+{x}_{8} {x}_{14}+{x}_{5} {x}_{15}-{x}_{6} {x}_{16}+1,-{x}_{11} {x}_{13}+{x}_{12} {x}_{14}+{x}_{9} {x}_{15}-{x}_{10} {x}_{16},0$
Edit
What @Youngsu suggests in a comment, does seem to work:
$\langle x_{12}+x_{15},x_{11}-x_{16},x_{10}+x_{13},x_9-x_{14},x_8+x_{14},x_7-x_{13},x_6-x_{16},x_5+x_{15},x_4+x_{13},x_3+x_{14},x_2-x_{15},x_1-x_{16},x_{13}^2+x_{14}^2+x_{15}^2+x_{16}^2-1\rangle$