Hint: First observe that
xCr+xCr+1=x+1Cr+1
and xCr+1+xCr+2=x+1Cr+2.
Now applying C3+C2 and C2+C1, we get
Δ=∣∣
∣
∣∣xCrx+1Cr+1x+1Cr+2yCry+1Cr+1y+1Cr+2zCrz+1Cr+1z+1Cr+2∣∣
∣
∣∣
Again applying C3+C2, we get
Δ=∣∣
∣
∣∣xCrx+1Cr+1x+2Cr+2yCry+1Cr+1y+2Cr+2zCrz+1Cr+1z+2Cr+2∣∣
∣
∣∣
by the same rule.