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Show that the four points having position vectors 6{\bar  {i}}-7{\bar  {j}},16{\bar  {i}}-19{\bar  {j}}-4{\bar  {k}},3{\bar  {i}}-6{\bar  {k}},2{\bar  {i}}+5{\bar  {j}}+10{\bar  {k}}\, are not coplanar.

Let A=6{\bar  {i}}-7{\bar  {j}},B=16{\bar  {i}}-19{\bar  {j}}-4{\bar  {k}},C=3{\bar  {i}}-6{\bar  {k}},D=2{\bar  {i}}+5{\bar  {j}}+10{\bar  {k}}\, be the given vectors.

Then {\bar  {AB}}=10{\bar  {i}}-12{\bar  {j}}-4{\bar  {k}},{\bar  {AD}}=-4{\bar  {i}}+12{\bar  {j}}+10{\bar  {k}},{\bar  {AC}}=-6{\bar  {i}}+10{\bar  {j}}-6{\bar  {k}}\,

Now [{\bar  {AB}}{\bar  {AC}}{\bar  {AD}}]={\begin{vmatrix}10&-12&-4\\-6&10&-6\\-4&12&10\end{vmatrix}}=10(100+72)+12(-60-24)-4(-72+40)=1720-1008+128\, which is not equal to zero.

Hence the vectors are not coplanar.

Hence the points A,B,C,D are not coplanar.

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