The line that is orthogonal to the plane 2x + 4y − 2z = 10 is x = -2t, y = t, z = 5t + 5.

To find this *line*, we can use the fact that a line that is orthogonal to a plane must be perpendicular to the normal vector of the plane. The normal vector of the plane 2x + 4y − 2z = 10 is (2, 4, -2), so we can find a point on the line that is perpendicular to this vector by setting the dot product of the point and the normal vector equal to zero.

If we let the point on the line be represented by (x, y, z), then we can write the equation as follows:

(2)(x) + (4)(y) + (-2)(z) = 0

Solving for x, y, and z, we get:

x = -2t y = t z = 5t + 5

where t is a real number that can be used to parameterize the line. This line is orthogonal to the plane 2x + 4y − 2z = 10.