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Q.
Let $A(0,0,0), B(1,1,1), C(3,2,1)$ and $D(3,1,2)$ be four points. The angle between the planes through the points $A, B, C$ and through the points $A, B, D$ is
Vector Algebra
Solution:
Let $n_{1}$ and $n_{2}$ be the vectors normal to the planes $A B C$ and $A B D$ respectively.
$n_{1}=A B \times A C=-i+2 j-k$
$n_{2}=A B \times A D=i+j-2 k$
Let $\theta$ be the acute angle between the planes, then $\theta$ is the acute angle between their normals $n_{1}$ and $n_{2}$
$\therefore \cos \theta=\frac{|-1+2+2|}{\sqrt{6} \cdot \sqrt{6}}$
$=\frac{3}{2}=\frac{1}{2}=\cos \frac{\pi}{3} $
$\Rightarrow =\frac{\pi}{3}$