UPSC NDA Math Model Paper 2

Here, we have to find the equation of the plane passing through the points A (3, 1, 2), B (6, 1, 2) and C (0, 2, 0) Here, $x_{1}=3, y_{1}=1, z_{1}=2, x_{2}=6, y_{2}=1,$ $z_{2}=2, x_{3}=0, y_{3}=2$ and $z_{3}=0$ As we know that, equation of the plane in Cartesian form 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 given by: $\begin{vmatrix} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{vmatrix}$ $=0$ $\Rightarrow \begin{vmatrix} x-3 & y-1 & z-2 \\ 3 & 0 & 0 \\ -3 & 1 & -2 \end{vmatrix}$ $=0$ $\Rightarrow (x - 3) \times (0 - 0) - (y - 1) \times (-6 - 0)$ $+ (z - 2) \times (3 - 0) = 0$ $\Rightarrow 6y - 6 + 3z - 6 = 0$ $\Rightarrow 6y + 3z - 12 = 0$ So, the equation of the required plane is 6y + 3z - 12 = 0 Hence, option D is the correct answer.