Q63 / 180 · KCET 2023 · Mathematics

The equation of the plane through the points $(2,1,0),(3,2,-2)$ and $(3,1,7)$ is

  • A
    $2 x-3 y+4 z-27=0$
  • B
    $6 x-3 y+2 z-7=0$
  • Correct answer
    C
    $7 x-9 y-z-5=0$
  • D
    $3 x-2 y+6 z-27=0$

Solution

We know that the equation of a plane passing through three non-collinear points. $\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right)$ and $\left(x_3, y_3, z_3\right)$ $\because \quad\left|\begin{array}{lll}x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1\end{array}\right|=0$ $\begin{aligned} & \Rightarrow \quad\left|\begin{array}{ccc}x-2 & y-1 & z-0 \\ 3-2 & 2-1 & -2-0 \\ 3-2 & 1-1 & 7-0\end{array}\right|=0 \\ & \Rightarrow \quad\left|\begin{array}{ccc}x-2 & y-1 & z \\ 1 & 1 & -2 \\ 1 & 0 & 7\end{array}\right|=0 \\ & \Rightarrow \quad(x-2)(7-0)-(y-1)(7+2)+z(0-1)=0 \\ & \Rightarrow \quad 7 x-14-(y-1) 9+(-z)=0 \\ & \Rightarrow \quad 7 x-9 y-z-5=0\end{aligned}$

Hence, the correct answer is option: 'C'.

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