If the product of distances of the point `(1,1,1)` from the origin and plane `x-y+z+lambda=0` be 5 then `lambda=`

Explanation

The distance of the point `(1,1,1)` from the origin


We know, distance of `(x_{1}, y_{1}, z_{1})` form the origin is :

`sqrt{{x}_{1}^{2}+y_{1}^{2}+z_{1}^{2}}`

Putting values of `x_{1}, y_{1}, z_{1}=1`

`Rightarrow` Required Distance `=sqrt{3}`

Distance of the point `(1,1,1)` from plane `x-y+z+lambda=0`

We know, the distance of point `(x_{1}, y_{1}, z_{1})` from the plane

`pi: a x+b y+c z+d=0` is given by:

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