Here →a1=−3^i+^j+5^k,→b1=−3^i+^j+5^k,→a2=−^i+2^j+5^k,→b2=−^i+2^j+5^k
Now (→a2−→a1).(→b1×→b2)=∣∣
∣∣210−315−125∣∣
∣∣=2(−5)−1(−15+5)=−10+10=0
∴ Given lines are coplanar.
Perpendicular vector to the plane →b1×→b2=∣∣
∣
∣∣^i^j^k−315−125∣∣
∣
∣∣=−5^i+10^j−5^k or ^i−2^j+^k
∴ Eq.of plane:→r.(^i−2^j+^k)=(^i−2^j+^k).(−3^i+^j+5^k) ⇒ x−2y+z=0