OABC is a tetrahedron where O is the origin and A,B,C have position vectors `veca,vecb,vecc` respectively prove that circumcentre of tetrahedron OABC is `(a^2(vecbxxvecc)+b^2(veccxxveca)+c^2(vecaxxvecb))/(2[veca vecb vecc])`

Explanation

To prove that the circumcenter of the tetrahedron OABC is given by the formula

\[ \vec{R} = \frac{a^2(\vec{b} \times \vec{c}) + b^2(\vec{c} \times \vec{a}) + c^2(\vec{a} \times \vec{b})}{2[\vec{a}, \vec{b}, \vec{c}]} \]

where \( \vec{a}, \vec{b}, \vec{c} \) are the position vectors of points A, B, and C respectively, we will follow these steps:

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