Q47 / 155 · WBJEE 2022 · Mathematics

The line $x-2 y+4 z+4=0, x+y+z-8=0$ intersect the plane $x-y+2 z+1=0$ at the point

  • A
    $(-2,5,1)$
  • B
    $(2,-5,1)$
  • C
    $(2,5,-1)$
  • Correct answer
    D
    $(2,5,1)$

Solution

Solving the \(3 \mathrm{eq}^n\) s will give as the required point of intersection \(\begin{aligned} & x-2 y+4 z=-4 \\ & x+y+z=8 \\ & x-y+2 z=-1 \\ & D=\left|\begin{array}{ccc} 1 & -2 & 4 \\ 1 & 1 & 1 \\ 1 & -1 & 2 \end{array}\right|=1(3)+2(1)+4(-2)=-3 \\ & \begin{aligned} D_1=\left|\begin{array}{rrr} -4 & -2 & 4 \\ 8 & 1 & 1 \\ -1 & -1 & 2 \end{array}\right|= & -4(3)+2(17)+4(-7) \\ & =-6 \end{aligned} \\ & D_2=\left|\begin{array}{ccc} 1 & -4 & 4 \\ 1 & 8 & 1 \\ 1 & -1 & 2 \end{array}\right|=-15 \\ & D_3=\left|\begin{array}{ccc} 1 & -2 & -4 \\ 1 & 1 & 8 \\ 1 & -1 & -1 \end{array}\right|=-3 \\ & x=\frac{D_1}{D}=2, \quad y=\frac{D_2}{D}=5, \quad z=\frac{D_3}{D}=1 \\ & \end{aligned}\) \((2, S, 1)\) is the required point.

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

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