Q42 / 155 · WBJEE 2024 · Mathematics

The plane $2 x-y+3 z+5=0$ is rotated through $90^{\circ}$ about its line of intersection with the plane $x+y+z=1$. The equation of the plane in new position is

  • A
    $3 x+9 y+z+17=0$
  • Correct answer
    B
    $3 x+9 y+z=17$
  • C
    $3 x-9 y-z=17$
  • D
    $3 x+9 y-z=17$

Solution

Hint : Equation of a plane passing through the line of intersection of the given planes is $2 x-y+3 z+5+\lambda(x+y+z-1)=0$ $\Rightarrow(2+\lambda) x-(1-\lambda) y+(3+\lambda) z+5-\lambda=0$ this will be perpendicular to the plane $2 x-y+3 z+5=0$ $\Rightarrow(2+\lambda) 2+(1-\lambda) \times 1+3(3+\lambda)=0 \Rightarrow \lambda=\frac{-7}{2}$ $\therefore$ Equation of plane $3 x+9 y+z=17$

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

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