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.