The plane $$x + 2y - z = 4$$ cuts the sphere $$x^2 + y^2 + z^2 - x + z - 2 = 0$$ in a circle of radius

The sphere and the plane intersect in a circle whose radius is obtained from
$$r^2 = R^2 - d^2,$$
where:

• $$R$$ is the radius of the sphere.
• $$d$$ is the perpendicular distance from the centre of the sphere to the plane.

Step 1: Write the sphere in centre-radius form.

Given sphere: $$x^2 + y^2 + z^2 - x + z - 2 = 0$$

Complete the squares for $$x$$ and $$z$$:

$$x^2 - x = (x - \tfrac12)^2 - \tfrac14,$$
$$z^2 + z = (z + \tfrac12)^2 - \tfrac14.$$

Substituting back,

$$(x-\tfrac12)^2 + y^2 + (z+\tfrac12)^2 - \tfrac14 - \tfrac14 - 2 = 0,$$
$$\Rightarrow (x-\tfrac12)^2 + y^2 + (z+\tfrac12)^2 = 2 + \tfrac14 + \tfrac14 = \tfrac52.$$

Hence centre $$C\!\left(\tfrac12,\,0,\,-\tfrac12\right)$$ and sphere radius
$$R = \sqrt{\tfrac52}.$$

Step 2: Find the distance from the centre to the plane.

Plane: $$x + 2y - z = 4.$$
Distance formula: $$d = \frac{|ax_0 + by_0 + cz_0 - 4|}{\sqrt{a^2 + b^2 + c^2}}$$ for point $$(x_0,y_0,z_0).$$

Here $$(a,b,c) = (1,2,-1)$$ and $$C = \left(\tfrac12,0,-\tfrac12\right):$$

$$d = \frac{\left|\tfrac12 + 2(0) - (-\tfrac12) - 4\right|}{\sqrt{1^2 + 2^2 + (-1)^2}} = \frac{|1 - 4|}{\sqrt6} = \frac3{\sqrt6} = \frac{\sqrt6}{2}.$$

Step 3: Compute the circle’s radius.

Square the quantities:
$$R^2 = \tfrac52,\qquad d^2 = \left(\frac{\sqrt6}{2}\right)^2 = \tfrac32.$$

Therefore
$$r^2 = R^2 - d^2 = \tfrac52 - \tfrac32 = 1 \;\;\Longrightarrow\;\; r = 1.$$

Hence the radius of the circle is $$1.$$

Option B which is: $$1$$