A tetrahedron has vertices O(0,0,0), A(1,2,1) | vedclass

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(B) The equation of the plane $OAB$ is given by the determinant form:$\begin{vmatrix} x-0 & y-0 & z-0 \ 1-0 & 2-0 & 1-0 \ 2-0 & 1-0 & 3-0 \end{vmatr

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$\frac{19}{35}$

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(B) The equation of the plane $OAB$ is given by the determinant form:$\begin{vmatrix} x-0 & y-0 & z-0 \ 1-0 & 2-0 & 1-0 \ 2-0 & 1-0 & 3-0 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} x & y & z \ 1 & 2 & 1 \ 2 & 1 & 3 \end{vmatrix} = 0$Expanding the determinant: $x(6-1) - y(3-2) + z(1-4) = 0 \Rightarrow 5x - y - 3z = 0$ ... $(i)$The equation of the plane $ABC$ is given by:$\begin{vmatrix} x-1 & y-2 & z-1 \ 2-1 & 1-2 & 3-1 \ -1-1 & 1-2 & 2-1 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} x-1 & y-2 & z-1 \ 1 & -1 & 2 \ -2 & -1 & 1 \end{vmatrix} = 0$Expanding the determinant: $(x-1)(-1+2) - (y-2)(1+4) + (z-1)(-1-2) = 0$$(x-1)(1) - (y-2)(5) + (z-1)(-3) = 0 \Rightarrow x - 1 - 5y + 10 - 3z + 3 = 0 \Rightarrow x - 5y - 3z + 12 = 0$ ... $(ii)$The angle $	heta$ between the planes $5x - y - 3z = 0$ and $x - 5y - 3z + 12 = 0$ is given by:$\cos 	heta = \frac{|(5)(1) + (-1)(-5) + (-3)(-3)|}{\sqrt{5^2 + (-1)^2 + (-3)^2} \sqrt{1^2 + (-5)^2 + (-3)^2}}$$\cos 	heta = \frac{|5 + 5 + 9|}{\sqrt{25 + 1 + 9} \sqrt{1 + 25 + 9}} = \frac{19}{\sqrt{35} \sqrt{35}} = \frac{19}{35}$

$A$ tetrahedron has vertices $O(0,0,0)$, $A(1,2,1)$, $B(2,1,3)$, and $C(-1,1,2)$. If $\theta$ is the angle between the faces $OAB$ and $ABC$, then $\cos \theta =$

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