AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

The equation of line joining the points.
a+2b−5c,−a−2b−3c is,
r=(a+2b−5c)+λ1(2a−4b−2c)
Similarly, cquation of line joining the points
−4c,6a−4b+4c is
r=(−4c)+λ2(6a−4b+8c)
Now, the point of intersection from above lines is.(2λ1+1)a+(4λ1+2)b+(−2λ−5)c
=(6λ2)a+(−4λ2)b+(8λ2+4)c
On comparing we get,
2λ1+1=6λ2
4λt+2=−4λ2
−2λ1−5=8λ2+4
On solving the above three equation we get,
λ1=−

1

2

and λ2=0
Thus, the intersection points are −4c and for µ=−3 the liner=(3a+6h−c)+µ(a+2b+c) passes through the point −4conly.