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Let the line L1: (x-1)/2=(y-2)/3=(z-3)/4...

Let the line `L_1: (x-1)/2=(y-2)/3=(z-3)/4` and `L_2:( x-2)/1= (y-3)/2=(z-k)/4` intersect at `P`.The least distance of `P` from the plane `3 x-4 y - 12 z+4=0`, equals

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Two lines (x-1)/(2)=(y-2)/(3)=(z-3)/(4) and (x-4)/(5)=(y-1)/(2)=(z)/(1) intersect at a point P. If the distance of P from the plane 2x-3y+6z=7 is lambda units, then the value of 49lambda is equal to

Knowledge Check

  • If the lines (x-1)/2=(y+1)/3=(z-1)/4 and (x-3)/1=(y-k)/2=z/1 , intersect, then

    A
    `2/9`
    B
    `9/2`
    C
    0
    D
    none of these
  • The point of intersection of the lines (x-1)/2=(y-2)/3=(z-3)/4 and (x-4)/5=(y-1)/2=z is

    A
    (0,0,0)
    B
    (1,1,1)
    C
    `(-1,-1,-1)`
    D
    (1,2,3)
  • If the lines (x-1)/(2)= (y+1)/(3)=(z-1)/(4) and (x-3)/(1)=(y-p)/(2)= (z)/(1) intersect, then p is equal to

    A
    `9/2`
    B
    10
    C
    1
    D
    `12/11`
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