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If the plane touches the sphere, then the perpendicular distance from the centre of the sphere to the plane should be equal to the radius of the given sphere.

$x^2+y^2+z^2-2x-4y+2z+3=0 \\=> (x-1)^2 + (y-2)^2 + (z+1)^2 = 9$

Thus centre of sphere is $(1,2,-1$ and radius is $3$.

Now, the distance from $(1,2,-1)$ to the plane is

$\frac{1.2 + 2.(-2) + (-1).1 + 12}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{9}{3} = 3$

which is equal to the radius of the sphere. Hence, the plane *touches *the sphere.

The equation of the line joining centre to the point of intersection can be written as:

$\frac{x-1}{2} = \frac{y-2}{-2} = \frac{z-1}{1} = r$

Any point of this line can be represented as $(2r+1, -2r+2, r-1)$

Putting this in the equation of the plane, we get $r = -1$. Hence, the point of intersection is $(-1, 4, -2)$

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