A tetrahedron has vertices P(1,2,1), Q(2,1,3) | vedclass

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(C) The equation of the plane $OPQ$ passing through $O(0,0,0), P(1,2,1)$,and $Q(2,1,3)$ is given by the determinant: $\begin{vmatrix} x & y & z \ 1 &

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$\cos ^{-1}\left(\frac{19}{35}\right)$

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(C) The equation of the plane $OPQ$ passing through $O(0,0,0), P(1,2,1)$,and $Q(2,1,3)$ is given by the determinant: $\begin{vmatrix} x & y & z \ 1 & 2 & 1 \ 2 & 1 & 3 \end{vmatrix} = 0$ $x(6-1) - y(3-2) + z(1-4) = 0 \Rightarrow 5x - y - 3z = 0$. The normal vector to plane $OPQ$ is $\vec{n_1} = 5\hat{i} - \hat{j} - 3\hat{k}$. The equation of the plane $PQR$ passing through $P(1,2,1), Q(2,1,3)$,and $R(-1,1,2)$ is given by: $\begin{vmatrix} x-1 & y-2 & z-1 \ 2-1 & 1-2 & 3-1 \ -1-1 & 1-2 & 2-1 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} x-1 & y-2 & z-1 \ 1 & -1 & 2 \ -2 & -1 & 1 \end{vmatrix} = 0$ $(x-1)(-1+2) - (y-2)(1+4) + (z-1)(-1-2) = 0$ $(x-1) - 5(y-2) - 3(z-1) = 0 \Rightarrow x - 5y - 3z + 12 = 0$. The normal vector to plane $PQR$ is $\vec{n_2} = \hat{i} - 5\hat{j} - 3\hat{k}$. The angle $	heta$ between the two planes is given by: $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} = \frac{|(5)(1) + (-1)(-5) + (-3)(-3)|}{\sqrt{5^2 + (-1)^2 + (-3)^2} \sqrt{1^2 + (-5)^2 + (-3)^2}}$ $\cos 	heta = \frac{|5 + 5 + 9|}{\sqrt{25+1+9} \sqrt{1+25+9}} = \frac{19}{\sqrt{35} \sqrt{35}} = \frac{19}{35}$ Therefore,$	heta = \cos^{-1}\left(\frac{19}{35}ight)$.

$A$ tetrahedron has vertices $P(1,2,1), Q(2,1,3), R(-1,1,2)$ and $O(0,0,0)$. Then the angle between the faces $OPQ$ and $PQR$ is

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