Finding equation of a circle I am supposed to find an equation of a circle passing though points M(3,0,0),N(0,3,0) and P(0,0,3) I have tried to find an intersection of bisectors of lines MN and NP but ended up with nothing.

It's not as hard as you may believe, in fact, you would use the following equation:
$$(x-a)^2+(y-b)^2+(z-c)^2=r^2$$
Just an altered version of the circle equation where $C(a,b,c)$ is the center of our circle.
So all you do is plug in you numbers to get 3 equations:
$$(3-a)^2+b^2+c^2=r^2\tag{1}$$
$$a^2+(3-b)^2+c^2=r^2\tag{2}$$
$$a^2+b^2+(3-c)^2=r^2\tag{3}$$
Subtract $(1)$ and $(2)$ to get:
$$(3-a)^2-a^2+b^2-(3-b)^2=0$$
$$9-6a-9+6b=0$$
$$6b=6a\to a=b$$
Subtract $(2)$ and $(3)$:
$$(3-b)^2-b^2+c^2-(3-c)^2=0$$
Following from above steps, we get:
$$a=b=c$$
From this, I want to assume that $a=b=c=0$ because I think it is obvious with the three points you chose and I want to have $r=3$, such that we have:
$$x^2+y^2+z^2=3^2=9$$
Plugging in your points say that this works and we have the following:
$$3=\pm\sqrt{x^2+y^2+z^2}=\pm\sqrt{x^2+(\sqrt{y^2+z^2})^2}$$
Basically, this is applying Pythagorean Theorem twice so that the points $x_1,y_1,z_1$ are all equidistant to the origin.
Since ask specifically for the circle, here you go:
$$(x-1)^2+(y-1)^2+(z-1)^2=6$$
Just remember to use the following equation:  $x+y+z=3$ to maintain the fact that it is a circle.
