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  • I have a doubt, kindly clarify. - Three Dimensional Geometry - JEE Main-3

Let P be the plane , which contains the line of intersection of the planes ,x+y+z-6=0\:\:and\:\:2x+3y+z+5=0,  and it is perpendicular to the xy-plane.  Then the distance of the point (0,0,256) from P is equal to :

  • Option 1)

    17/\sqrt{5}

  • Option 2)

    63\sqrt{5}

  • Option 3)

    205 \sqrt{5}

  • Option 4)

    11/\sqrt{5}

 

Answers (1)

best_answer

equation of plane which contains intersection of the planes x+y+z-6=0\:\:and\:\:2x+3y+z+5=0 is \left ( x+y+z-6 \right )+\lambda\left ( 2x+3y+z+5 \right )=0

\\x(1+2\lambda)+y(1+3\lambda)+2(1+\lambda)-6+5\lambda=0

since it is  \perp to xy plane

\\1+\lambda=0\\\\\:\lambda=-1

so  

\\x(1-2)+y(1-3)+z(0)-6+5(-1)=0\\\\\:-x-2y-11=0

x+2y+11=0 

distant of point (0,0,256) from this plane  =     \frac{11}{\sqrt{1+4}}=11/\sqrt{5}


Option 1)

17/\sqrt{5}

Option 2)

63\sqrt{5}

Option 3)

205 \sqrt{5}

Option 4)

11/\sqrt{5}

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