Let →A=−2^i+3^j+5^k, →B=^i+2^j+3^k and →C=7^i−^k.
Now, line −−→AB=→A+λ(→B−→A)
−−→AB=−2^i+3^j+5^k+λ(^i+2^j+3^k−(−2^i+3^j+5^k))
−−→AB=−2^i+3^j+5^k+λ(3^i−^j−2^k)
Now, if →C lies on −−→AB, we should have
7^i−^k=−2^i+3^j+5^k+λ(3^i−^j−2^k)
If λ=3, −−→AB=7^i−^k
OR
Let →A=−2^i+3^j+5^k, →B=^i+2^j+3^k and →C=7^i−^k.
Now, line −−→AB=→B−→A
∴−−→AB=^i+2^j+3^k−(−2^i+3^j+5^k)
∴−−→AB=3^i−^j−2^k
Now, −−→AC=→C−→A
∴−−→AC=7^i−^k−(−2^i+3^j+5^k)
∴−−→AC=9^i−3^j−6^k
Also, if −−→AB and −−→AC are collinear, −−→AB×−−→AC should be 0.
∴−−→AB×−−→AC=∣∣
∣
∣∣^i^j^k3−1−29−3−6∣∣
∣
∣∣
=(6−6)^i−(−18+18)^j+(−9+9)^k
=0