Find the vector equation of the plane passing through three points with position vectors ^i+^j-^k, 2^i-^j+^k and ^i+2^j+^k. Also find the coordinates of the point of intersection of this plane and the line →r=3^i-^j-^k+λ(2^i-^j+^k).

(→r−→a) (→b×→c)=0
→n=→b×→c=∣∣∣∣∣→i→j→k2−11121∣∣∣∣∣=7→i−→j+5→k
Therefore,[→r−(→i+→j−→k)] (7→i−→j+5→k)=0
Equation of plane is
→r.(7→i−→j+5→k)=14
Now, →r=3→i−→j−→k+λ(2→i−→j+→k)
This line intersects the plane,
Therefore,[(3+2λ)→i+(−1−λ)→j(−1+λ)→k].[7→i−→j+5→k]=14
21+14λ+1+λ−5+5λ=14
20λ=14−16=−2
λ=−1/10
Therefore, Point of intersection is
(3−2/10)→i+(−1+1/10)→j+(−1−1/10)→k=28/10→i−9/10→j−11/10→k