To find the shortest distance from a point to a surface, we need to find the perpendicular distance or the normal distance between them.

The surface of the sphere x^2 + y^2 + z^2 = 24 has center at the origin (0,0,0) and radius √24.

To find the normal distance from the point (1,2,-1) to the surface, we need to find the normal vector to the surface at the point on the surface that is closest to the point (1,2,-1). This point should lie on the line connecting the origin to the point (1,2,-1).

Let's find the intersection of this line with the surface of the sphere.

The line connecting the origin to the point (1,2,-1) can be parametrized as follows:

x = t(1) = t
y = t(2) = 2t
z = t(-1) = -t

Substituting these expressions into the equation of the sphere, we get:

t^2 + (2t)^2 + (-t)^2 = 24

Simplifying, we get:

6t^2 = 24

t^2 = 4

t = ± 2

So the two points of intersection are:

(2,4,-2) and (-2,-4,2)

We want the point on the surface that is closest to the point (1,2,-1), so we choose the point (2,4,-2).

The normal vector to the sphere at this point is simply the vector from the origin to (2,4,-2), which is:

<2,4,-2>

So the normal distance from the point (1,2,-1) to the surface of the sphere is given by the projection of the vector from (1,2,-1) to (2,4,-2) onto the normal vector to the sphere:

distance = |proj<1-2,2-4,-1-(-2)> · <2,4,-2>| / |<2,4,-2>|

= |<-1,-2,1> · <2,4,-2>| / √(2^2 + 4^2 + (-2)^2)

= |-2-8+2| / √24

= 12 / √24

= 3√6

So the shortest distance from the point (1,2,-1) to the surface of the sphere is 3√6.