Show that the points `P(a+2b+c),Q(a-b-c),R(3a+b+2c) and S(5a+3b+5c)` are coplanar given that a,b and c are non-coplanar.

Explanation

To show that P,Q,R,S are coplanar, we will show that PQ,PR,PS are coplanar
`PQ=-3b-2c`
`PR=2a-b+c`
`PS=4a+b+4c`
Let `PQ=xPR+yPS`
`implies-3b-2c=x(2a-b+c)+y(4a+b+4c)`
`implies-3b-2c=(2x+4y)a+(-x+y)b+(x+4y)c`
As the vectors a,b,c are non-coplanar, we can equate their coefficients.
`implies 0=2x+4y`
`implies-3=-x+y`
`implies-2=x+4y`
`x=2,y=-1` is unique solutionn for the above system of equations.
`impliesPQ=2PR-PS`
PQ,PR,PS are coplanar because PQ is a linear combination of PR and PS
`implies`The points P,Q,R,S are also coplanar.
trick for the vectors PQ,PR and PS to be coplanar, we
must have `|(0,-3,-2),(2,-1,1),(4,1,4)|=0` which is true
`therefore`The PQ,PR,PS are coplanar.
Hence, the points P,Q,R,S are also coplanar.