Three non - zero vectors `bar(a),bar(b),bar(c)` are complanar if and only if there exist scalars x,y,z, not all zero simultaneously such that `xbar(a)+ybar(b)+zbar(c)=bar(0)`.

Explanation

Let `bar(a),bar(b),bar(c)` be coplanar vectors. Then any one of them, say `bar(a)`, will be the linear combination of `bar(b)andbar(c)`.
`:.` there exist scalars `alphaandbeta` such that
`bar(a)=alphabar(b)+betabar(c)`
`:.(-1)bar(a)+alphabar(b)+betabar(c)=bar(0),i.e.,xbar(a)+ybar(b)+zbar(c)=bar(0)`
where `x=-1,y=alpha,z=beta` which are not all zero simultaneously.
Conversely : Let there exist scalars x,y,z not all zero such that
`xbar(a)+ybar(b)+zbar(c)=bar(0)` . . . (1)
Let `x!=0`, then divide (1) by s, we get
`i.e.,bar(a)+((y)/(x))bar(b)+((z)/(x))bar(c)=bar(0)`
`:.bar(a)=(-(y)/(x))bar(b)+(-(z)/(x))bar(c)`
`i.e.,bar(a)=alphabar(b)+betabar(c)`, where `alpha=(-y)/(x)andbeta=(-z)/(x)` are scalars.
`:.bar(a)` is the linear combination of `bar(b)andbar(c)`.
Hence, `bar(a),bar(b),bar(c)` are coplanar.