Given equations are
x+y+z=12x+Ny+2z=23x+3y+Nz=3
So,
Δ=∣1231N312N∣
Applying C2↔C2−C1 and C3↔C3−C1
⇒Δ=∣1230N−2000N−3∣
⇒Δ=(N−2)(N−3)
For unique solution Δ=0, so
N=2,3
Also, N∈1,2,3,4,5,6 and for unique solution N can take values 1,4,5,6.
⇒P(system has unique solution ) =64=6k
So, k=4
Therefore, required sum is
=4+1+4+5+6=20