Shift and Rotate

Since rotating the resulting points (after the shift) about a certain axis that passes through the origin results in $A'$ , $B'$ , and $C'$ , the new points after the shift must be the same distance away from the origin as $A'$ , $B'$ , and $C'$ , so:

$(5+d_x)^2+(6+d_y)^2+(10+d_z)^2=1^2+1^2+0^2$

$(5+d_x)^2+(18+d_y)^2+(26+d_z)^2=1^2+(-19)^2+0^2$

$(-16+d_x)^2+(6+d_y)^2+(10+d_z)^2=22^2+1^2+0^2$

which solves to $(d_x, d_y, d_z) = (-6, -\frac{33}{5}, -\frac{54}{5})$ , so $d_x + d_y + d_z = -\frac{117}{5} = \boxed{-23.4}$ .

Let $R$ be the rotation matrix. The following relations are described in the problem statement:

$R (A + d) = A' \\ R (B + d) = B' \\ R (C + d) = C'$

The rotation matrix is defined by an axis and an angle. Let the axis coincide with the unit vector whose components are:

$u_x = \cos \theta \, \sin \phi \\ u_y = \sin \theta \, \sin \phi \\ u_z = \cos \phi$

Let $\alpha$ be the rotation angle about the axis. The rotation matrix is defined as per the description at the Wikipedia page in the section "Rotation matrix from axis and angle". The equations are also given in the code below.

It is evident from the first set of equations that the rotation matrix effectively determines the $d$ vector.

$d = R^{-1} A' - A$

I used the following process in conjunction with a hill-climbing algorithm:

1) Move around randomly in $(\theta, \phi, \alpha)$ space via mutation
2) For each $(\theta, \phi, \alpha)$ , form a candidate rotation matrix $R$
3) From the candidate rotation matrix and the other information, calculate $d = R^{-1} A' - A$
4) Form a residual quantity. $\,\,\,\,\,\, \text{residual} = |R (B + d) - B'| + |R (C + d) - C'|$
5) If the most recent mutation results in the smallest residual thus far seen, store the mutated values. Otherwise, reject them
6) This process results in $(d_x,d_y,d_z) = (-6.0, -6.6, -10.8)$

Code

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import math
import random
import numpy as np

##################################################################

# Vector A,B,C components

Ax = 5.0
Ay = 6.0
Az = 10.0

Bx = 5.0
By = 18.0
Bz = 26.0

Cx = -16.0
Cy = 6.0
Cz = 10.0

##################################################################

# Vector A',B',C' components

Apx = 1.0
Apy = 1.0
Apz = 0.0

Bpx = 1.0
Bpy = -19.0
Bpz = 0.0

Cpx = 22.0
Cpy = 1.0
Cpz = 0.0

###################################################################

# Vectors A,B,C,A',B',C'

A = np.array([Ax,Ay,Az])
B = np.array([Bx,By,Bz])
C = np.array([Cx,Cy,Cz])

Ap = np.array([Apx,Apy,Apz])
Bp = np.array([Bpx,Bpy,Bpz])
Cp = np.array([Cpx,Cpy,Cpz])

###################################################################

delta = 0.001  # mutation step

minres = 99999999.0  # cumulative residual

theta = 2.0*math.pi*random.random()   # random parameter initialization
phi = math.pi*random.random()

alpha = 2.0*math.pi*random.random()

count = 0

for j in range(0,2*10**6):   # number of rounds of mutation

    thetam = theta + delta * (-1.0 + 2.0 * random.random())   # mutated values
    phim = phi + delta * (-1.0 + 2.0 * random.random())
    alpham = alpha + delta * (-1.0 + 2.0 * random.random())

    ########

    # unit vector used to form rotation matrix

    ux = math.cos(thetam) * math.sin(phim)
    uy = math.sin(thetam) * math.sin(phim)
    uz = math.cos(phim)

    ########

    # rotation matrix entries

    R11 = math.cos(alpham) + (ux**2.0)*(1.0 - math.cos(alpham))
    R12 = ux*uy*(1.0 - math.cos(alpham)) - uz*math.sin(alpham)
    R13 = ux*uz*(1.0 - math.cos(alpham)) + uy*math.sin(alpham)

    R21 = uy*ux*(1.0 - math.cos(alpham)) + uz*math.sin(alpham)
    R22 = math.cos(alpham) + (uy**2.0)*(1.0 - math.cos(alpham))
    R23 = uy*uz*(1.0 - math.cos(alpham)) - ux*math.sin(alpham)

    R31 = uz*ux*(1.0 - math.cos(alpham)) - uy*math.sin(alpham)
    R32 = uz*uy*(1.0 - math.cos(alpham)) + ux*math.sin(alpham)
    R33 = math.cos(alpham) + (uz**2.0)*(1.0 - math.cos(alpham))

    ########

    # rotation matrix and its inverse

    R =  np.array([[R11,R12,R13],[R21,R22,R23],[R31,R32,R33]])
    Rinv = np.linalg.inv(R)

    ########

    # calculate d vector based on candidate R matrix

    d = np.dot(Rinv,Ap) - A

    ########

    # Calculate residual components from vectors B and C

    Bcheck = np.dot(R,B+d) - Bp
    Ccheck = np.dot(R,C+d) - Cp

    ########

    # residual

    res = np.linalg.norm(Bcheck) + np.linalg.norm(Ccheck)

    ########

    # accept mutated values if cumulative residual gets smaller

    if res < minres:

        minres = res

        theta = thetam 
        phi = phim 
        alpha = alpham

        dx_store = d[0]
        dy_store = d[1]
        dz_store = d[2]

    count = count + 1

    if count%10**6 == 0:
        print count/10**6

##################################################################

print ""
print ""
print minres
print ""
print dx_store
print dy_store
print dz_store
print ""
print (dx_store+dy_store+dz_store)

print ""
print "#######################"
print ""

The images of $A, B, C$ after the shift and rotation are given by, $A' = R (A + d) , B' = R (B + d), C' = R(C + d)$ , where $d$ is the shift vector and $R$ is the rotation matrix. Substracting the first from the second, and the first from the third, results in:

$[ B' - A' \hspace{12pt} C' - A ' ] = R [ B - A \hspace{12pt} C - A ]$

To determine $R$ we need a third vector on either side of the equation, and for that we can use the cross product, so the equation becomes:

$[ B' - A' \hspace{12pt} C' - A' \hspace{12pt} (B' - A') \times (C' - A') ] = R [ B - A \hspace{12pt} C - A \hspace{12pt} (B-A) \times (C - A)]$

And this is of the form, $U = R V$ , so we can determine $R$ by matrix inversion then multiplication, namely, $R = U V^{-1}$ .

Now that $R$ is determined, we use any of the first three equations, to determine the vector $d$ . For example, since $A' = R (A + d)$ , it follows that $d = -A + R^T A'$ .

Carrying out the aforementioned operations, results in $d = [-6, -6.6, -10.8]^T$ ; therefore, $(d_x + d_y + d_z) = \boxed{-23.4}$ .