Find the equation of the plane through the points (2,1,0), (3,-2,-2) and (3,1,7).

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Solution

We know that, the equation of a plane passing through three non-collinear points $(x_{1}, y_{1}, z_{1})$ , $(x_{2}, y_{2}, z_{2})$ and $(x_{3}, y_{3}, z_{3})$ is $∣ ∣ ∣ ∣ \begin{matrix} x - x_{1} & y - y_{1} & z - z_{1} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} x_{3} - x_{1} & y_{3} - y_{1} & z_{3} - z_{1} \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} x - 2 & y - 1 & z - 0 3 - 2 & - 2 - 1 & - 2 - 0 3 - 2 & 1 - 1 & 7 - 0 \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} x - 2 & y - 1 & z 1 & - 3 & - 2 1 & 0 & 7 \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow (x - 2) (- 21 + 0) - (y - 1) (7 + 2) + z (3) = 0 \Rightarrow - 21 x + 42 - 9 y + 9 + 3 z = 0 \Rightarrow - 21 x + 9 y + 3 z = - 51 ∴ 7 x - 3 y - z = 17$
So, the required equation of plane is $7 x + 3 y - z = 17.$

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