→r=→a+λ(→b−→a) (equation of a line)
So, the 2 intersecting lines are,
→r1=(2a+3b−c)+λ((3a+4b−2c)−(2a+3b−c))=(2a+3b−c)+λ(a+b−c)
and →r2=(a−2b+3c)+μ((a−6b+6c)−(a−2b+3c))=(a−2b+3c)+μ(−4b+3c)
At the point of intersection, →r1=→r2.
ie, (2a+3b−c)+λ(a+b−c)=(a−2b+3c)+μ(−4b+3c)
As a,b,c are non-coplanar, there could only be one linear combination of the vectors for each spanning vector. (characteristic of vector space)
ie, the coefficients of a,b,c are unique.
From equation of →r2 we can see that whatever the value of μ, the coefficient of a is constant and is 1.
But for →r1 only λ=(−1) gives coefficient 1 for a.
So, the intersection point →r=(2a+3b−c)+(−1)(a+b−c)=(a+2b)