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Let $\Pi_1=\operatorname{span}\{e_1,e_2\},\Pi_2=\operatorname{span}\{e_3,e_4\},\Pi_3=\operatorname{span}\{e_5,e_6\}$ be orthogonal 2D planes in $\mathbb{R}^6$.

Let $U$ be arbitrary 2D plane in $\mathbb{R}^6$.

Let $P_i:U\to\Pi_i\;(i=1,2,3)$ be the orthogonal projections.

Let $U_{ij}\;(i=1,2,3;j=1,2,3;i<j)$ be the orthogonal projection of $U$ to $\Pi_i\oplus\Pi_j$. So $U_{ij}$ are 2D planes.

Let $P_{ij}:U\to U_{ij}\;(i=1,2,3;j=1,2,3;i<j)$ be the orthogonal projections.

Then I think the following is true: $$1-(\det P_{12})^2-(\det P_{13})^2-(\det P_{23})^2\\+(\det P_1)^2+(\det P_2)^2+(\det P_3)^2=0$$ I proved it with heavy computation below. Is there a better proof? Can the identity be generalized? Thanks.

We can expand to verify the polynomial identity: \begin{align}\tag1 0&=(a_1^2+\dots+a_6^2)(b_1^2+\dots+b_6^2)-(a_1 b_1+\dots+a_6 b_6)^2\\ &-\left(\left(a_1^2+a_2^2+a_3^2+a_4^2\right) \left(b_1^2+b_2^2+b_3^2+b_4^2\right)-(a_1 b_1+a_2 b_2+a_3 b_3+a_4 b_4)^2\right)\\ &-\left(\left(a_1^2+a_2^2+a_5^2+a_6^2\right) \left(b_1^2+b_2^2+b_5^2+b_6^2\right)-(a_1 b_1+a_2 b_2+a_5 b_5+a_6 b_6)^2\right)\\ &-\left(\left(a_3^2+a_4^2+a_5^2+a_6^2\right)\left(b_3^2+b_4^2+b_5^2+b_6^2\right)-(a_3 b_3+a_4 b_4+a_5 b_5+a_6 b_6)^2\right)\\ &+\left(\left(a_1^2+a_2^2\right) \left(b_1^2+b_2^2\right)-(a_1 b_1+a_2 b_2)^2\right)\\ &+\left(\left(a_3^2+a_4^2\right) \left(b_3^2+b_4^2\right)-(a_3 b_3+a_4 b_4)^2\right)\\ &+\left(\left(a_5^2+a_6^2\right) \left(b_5^2+b_6^2\right)-(a_5 b_5+a_6 b_6)^2\right) \end{align} Let $\{(a_1,\dots,a_6),(b_1,\dots,b_6)\}$ be an orthonormal basis for $U$, then $$P_1((a_1,a_2,a_3,a_4,a_5,a_6))=(a_1,a_2,0,0,0,0)\\ P_{12}((a_1,a_2,a_3,a_4,a_5,a_6))=(a_1,a_2,a_3,a_4,0,0)$$ so the identity $(1)$ becomes, \begin{align}0&=1-0\\ &-\det P_{12}\\ &-\det P_{13}\\ &-\det P_{23}\\ &+\det P_1\\ &+\det P_2\\ &+\det P_3\end{align}

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