Calculations:

We calculate the triple product of the three given vectors in order to find the volume of the parallelogram they are shaping, the triple product of the three vectors can be calculated for the vectors in any differen order, we start by assigning the vectors in the determinant as follows

\(c\cdot(a\times b)=\begin{vmatrix}2&1&4\\1&2&3\\-1&1&2\end{vmatrix}\)

\(\triangle=\begin{vmatrix}2&1&4\\1&2&3\\-1&1&2\end{vmatrix}\begin{vmatrix}2&1\\1&2\\-1&1\end{vmatrix}\)

We then, calculate the value \(\triangle\) of the determinant, using Leibniz or Sarrus, using Sarrus's. The value of the derminant according to Sarrus's the sum of the South-East diagonals, minus the South-West diagonals.

\(=(((2\times2\times2)+(1\times3\times-1)+(4\times1\times1))-((1\times1\times2)+(2\times3\times1)+(4\times2\times-1)))\)

\(=((8-3+4)-(2+6-8))\)

\(=((9)-(0))\)

\(=9\) Thus, the value of the determinant is 9, if we changed the order of the vectors in the triple product the value would be either -9 or +9, in either case we take the absolute value.

\([c\cdot(a\times b)]=|9|\)

\(=9\)

Thus, the volume of the parallelogram shaped by the three vectors a, b and c is 9 units cubed, where the referring unit is the unit of the distance between the two points or the unit of the scale in general by which the points were coordinated.

Result: 9 units

We calculate the triple product of the three given vectors in order to find the volume of the parallelogram they are shaping, the triple product of the three vectors can be calculated for the vectors in any differen order, we start by assigning the vectors in the determinant as follows

\(c\cdot(a\times b)=\begin{vmatrix}2&1&4\\1&2&3\\-1&1&2\end{vmatrix}\)

\(\triangle=\begin{vmatrix}2&1&4\\1&2&3\\-1&1&2\end{vmatrix}\begin{vmatrix}2&1\\1&2\\-1&1\end{vmatrix}\)

We then, calculate the value \(\triangle\) of the determinant, using Leibniz or Sarrus, using Sarrus's. The value of the derminant according to Sarrus's the sum of the South-East diagonals, minus the South-West diagonals.

\(=(((2\times2\times2)+(1\times3\times-1)+(4\times1\times1))-((1\times1\times2)+(2\times3\times1)+(4\times2\times-1)))\)

\(=((8-3+4)-(2+6-8))\)

\(=((9)-(0))\)

\(=9\) Thus, the value of the determinant is 9, if we changed the order of the vectors in the triple product the value would be either -9 or +9, in either case we take the absolute value.

\([c\cdot(a\times b)]=|9|\)

\(=9\)

Thus, the volume of the parallelogram shaped by the three vectors a, b and c is 9 units cubed, where the referring unit is the unit of the distance between the two points or the unit of the scale in general by which the points were coordinated.

Result: 9 units