JEE Main 2019 · 9 Apr · Shift II

JEE Main 2019 Mathematics question on Three Dimensional Geometry

Let PP be the plane, which contains the line of intersection of the planes, x+y+z−6=0x + y + z - 6 = 0 and 2x+3y+z+5=02x + 3y + z + 5 = 0 and it is perpendicular to the XYXY-plane. Then, the distance of the point (0,0,256)(0, 0, 256) from PP is equal to
  1. A.
    63563\sqrt{5}
  2. B.
    2055205\sqrt{5}
  3. C.
    115\dfrac{11}{\sqrt{5}}
  4. D.
    175\dfrac{17}{\sqrt{5}}
Show answer

Correct option: C

[JEE Main 2019 · 9 Apr · Shift II]

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