If the point of intersection of the plan...

If the point of intersection of the plane `4x - 5y + 2z - 6 = 0` with the line through the origin and perpendicular to the plane `x- 2y - 4z = 4` is P, then the distance of the point P from (1, 2,3) is

`sqrt(63)` units

8 units

`sqrt(65)` units

`sqrt(72)` units

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The plane x + 3y + 13 = 0 passes through the line of intersection of the planes 2x - 8y + 4z = p and 3x - 5y + 4z + 10 =0 . If the plane is perpendicular to the plane 3x - y - 2z - 4 = 0 , then the value of p is equal to

The plane x + 3y + 13 = 0 passes through the line of intersection of the planes 2x - 8y + 4z = p and 3x - 5y + 4z + 10 = 0 If the plane is perpendicular to the plane , 3x - y - 2z - 4 = 0 then the value of p is equal to

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Find the equation of the plane through the line of intersection of the planes x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x-y+z=0 Then find the distance of plane thus obtained from the point A(1,3,6) .

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