Q52 / 90 · JEE Main 2019 (11 Jan Shift 2) · Mathematics

If the point $(2, \alpha, \beta)$ lies on the plane which passes through the points (3,4,2) and (7,0,6) and is perpendicular to the plane $2 x-5 y=15,$ then $2 \alpha-3 \beta$ is equal to :

  • A
    12
  • Correct answer
    B
    7
  • C
    5
  • D
    17

Solution

Let the normal to the required plane is $\vec{n}$, then $\vec{n}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 4 & -4 & 4 \\ 2 & -5 & 0\end{array}\right|=20 \hat{i}+8 \hat{j}-12$ $\therefore \quad$ Equation of the plane $(x-3) \times 20+(y-4) \times 8+(z-2) \times(-12)=0$ $5 x-15+2 y-8-3 z+6=0$ $5 x+2 y-3 z-17=0 \quad \ldots(1)$ Since, equation of plane (1) passes through $(2, \alpha, \beta)$, then $10+2 \alpha-3 \beta-17=0 \Rightarrow 2 \alpha-3 \beta=7$

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

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