Given, the system of equations
x+y+z=6
x+2y+αz=10
x+3y+5z=β
Now finding,
△=∣1111231α5∣
⇒△=1(10−3α)−(5−α)+(3−2)
⇒△=6−2α
Now for unique solution, △=0⇒α=3
So, for unique solution ⇒α=3
But in option (D), system has unique solution for α=3 which is completely wrong.