The correct option is A μ, λ2, v are in A.P.
We have,
→r=3^i+2^j−5^k⋯(i)
and →r=λ→a+μ→b+v→c
⇒→r=λ(2^i−^j+^k)+μ(^i+3^j−2^k)+v(−2^i+^j−3^k)
⇒→r=^i(2λ+μ−2v)+^j(−λ+3μ+v)+^k(λ−2μ−3v)⋯(ii)
Comparing like vectors, we get
2λ+μ−2v=3⋯(iii)
−λ+3μ+v=2⋯(iv)
λ−2μ−3v=−5⋯(v)
Solving these above equations, we get
μ=1, v=2, λ=3
Hence, μ, λ2, v are in A.P.