# 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) $11/\sqrt{5}$

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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