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I am trying to simplify this triple product $$(2\vec a + \vec b - \vec c,\vec b,\vec a - \vec c)$$

My result is: $$(\vec a,\vec b, -\vec c)$$ There is no such possible solution between four given answers. What am I doing wrong?

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    What are the options? – 2017-02-07
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    @pepa.dvorak yes – 2017-02-07
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    What are those four given answers? What about $-(a,b,c)$? – 2017-02-07
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    @pepa.dvorak a) 2(a,b,c), b) 2(a,c,b), c) (a,b,c), d) (a,c,b) – 2017-02-07

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You've arrived at a good point with $(a,b,-c)$. Now just use properties of determinant: switching two lines/columns changes the sign of the determinant, so $(a,b,-c) = -(a,-c,b)$ and multiplying a line/column by a constant $c$ multiplies the determinant $c$ times, hence $-(a,-c,b)= (a,c,b)$.

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    I understand the first part, but: "multiplying a line/column by a constant c multiplies the determinant c times, hence −(a,−c,b)=(a,c,b)" I do not. Can you explain more in details? – 2017-02-07
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    Since $det(AB) = det(A)\cdot det(B)$, take $B= \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ and $A= (a, -c, b)$. Then $AB = (a,c,b)$ and $det(AB) = - det (A) = $ "triple product of $(a, -c, b)$". – 2017-02-07