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  4. 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 :-

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Q. 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 :-

JEE MainJEE Main 2019Three Dimensional Geometry

Solution:

$\lambda\left(x+y+z-6\right) +2x+3y+z+5=0 $
$\left(\lambda+2\right)x+\left(\lambda+3\right)y+\left(\lambda+1\right)z+5 -6\lambda=0 $
$ \lambda+1 =0 \Rightarrow \lambda=-1 $
$ P:x +2y +11 =0 =\frac{11}{\sqrt{5}} $