If ` vec aa n d vec b` are two non-collinear vectors, show that points `l_1 vec a+m_1 vec b ,l_2 vec a+m_2 vec b` and `l_3 vec a+m_3 vec b` are collinear if `|l_1l_2l_3m_1m_2m_3 1 1 1|=0.`

Explanation

We know that three points having P.V.s `veca, vecb and vecc` are collinear if there exists a relation of the form
`xveca+yvecb+zvecc, =vec0`, where `x+y+z=0`.
Now `xveca+yvecb+zvecc=vec0` gives
`" "x(l_(1)veca+m_(1)vecb)+y(l_(2)veca+m_(2)vecb)+z(l_(3)veca+m_(3)vecb)=vec0`
or `" "(xl_(1)+yl_(2)+zl_(3))veca+(xm_(1)+ym_(2)+zm_(3))vecb=vec0`
Since `veca and vecb` are two non-collinear vectors, it follows that
`" "xl_(1)+yl_(2)+zl_(3)=0" "` (i)
`" "xm_(1)+ym_(2)+zm_(3)=0" "`(ii)
Because otherwise one is expressible as a scalar multiple of the other which would mean that `veca and vecb` are collinear. Also
`" "x+y+z=0" "`(iii)
Eliminating `x, y and z` from (i), (ii) and (iii), we get
`" "|{:(l_(1),,l_(2),,l_(3)),(m_(1),,m_(2),,m_(3)),(1,,1,,1):}|=0`
Alternate method :
`" "A(l_(1)veca+m_(1)vecb), B(l_(2)veca+m_(2)vecb) and C(l_(3)veca+m_(3)vecb)` are collinear.
`rArr" "` Vectors = `(l_(2)-l_(3))veca+(m_(2)-m_(3))vecb and vec(AB) =(l_(1) -l_(2))veca+(m_(1)-m_(2))vecb` are collinear.
`rArr" "(l_(1)-l_(2))/(l_(2)-l_(3))=(m_(1)-m_(2))/(m_(2)-m_(3))`
`rArr" "|{:(l_(1),,l_(2),,l_(3)),(m_(1),,m_(2),,m_(3)),(1,,1,,1):}|=0`