Given,
a=6i^+9j^+12k^,b=αi^+11j^−2k^ and c be vectors such that a×c=a×b,
a×c−a×b=0
⇒a×(c−b)=0
So, \vec{a}&(\vec{c}-\vec{b}) are parallel vectors,
Hence, λa=c−b
⇒c=b+λa
⇒a⋅c=a⋅b+λ∣a∣2
⇒−12=(6α+75)+λ(261)
⇒2α+87λ=−29....(i)
Now again using c=b+λa we get,
c=i^(α+6λ)+j^(11+9λ)+k^(−2+12λ)
Also given c⋅(i^−2j^+k^)=5
⇒(α+6λ)−2(11+9λ)+(−2+12λ)=5
⇒α=29
So, 2α+87λ=−29
⇒λ=−1
Hence, c=23i^+2j^−14k^
So, the value of c⋅(i^+j^+k^)=23+2−14=11