find the value of p so that the points with position vector 6i-j ,16i-29j-4k ,3j-6k ,2i+5j+pk are coplanar

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Solution

Dear student,

Four points are coplanar if and only if the volume of the tetrahedron defined by them is 0.
i.e. $| \begin{matrix} x_{1} & y_{1} & z_{1} & 1 \\ x_{2} & y_{2} & z_{2} & 1 \\ x_{3} & y_{3} & z_{3} & 1 \\ x_{4} & y_{4} & z_{4} & 1 \end{matrix} | = 0$
Now, according to the question:
$(x_{1}, y_{1}, z_{1}) = (6, - 1, 0) (x_{2}, y_{2}, z_{2}) = (16, - 29, - 4) (x_{3}, y_{3}, z_{3}) = (0, 3, - 6) (x_{4}, y_{4}, z_{4}) = (2, 5, p) H e n c e, | \begin{matrix} 6 & - 1 & 0 & 1 \\ 16 & - 29 & - 4 & 1 \\ 0 & 3 & - 6 & 1 \\ 2 & 5 & p & 1 \end{matrix} | = 0 o r, 6 |\begin{matrix} - 29 & - 4 & 1 \\ 3 & - 6 & 1 \\ 5 & p & 1 \end{matrix}| + |\begin{matrix} 16 & - 4 & 1 \\ 0 & - 6 & 1 \\ 2 & 5 & p \end{matrix}| - |\begin{matrix} 16 & - 29 & - 4 \\ 0 & 3 & - 6 \\ 2 & 5 & p \end{matrix}| = 0 o r, 6 (8 p + 49) - 4 (4 p + 23) - 48 p - 852 = 0 o r, 48 p + 294 - 16 p - 92 - 48 p - 852 = 0 o r, - 16 p - 650 = 0 o r, - 16 p = 650 p = - \frac{650}{16} = - 40.625$

Regards

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