The volume of the parallelepiped is 55 cubic units.
In the question, we are asked to find the volume of the parallelepiped with adjacent edges PQ, PR, and PS, given P(−2, 1, 0), Q(5, 3, 5), R(1, 4, −1), S(3, 6, 2).
We first find the vectors:
PQ = Q - P
= < 5 - (-2), 3 - 1, 5 - 0 >
= < 7, 2, 5 >.
PR = R - P
= < 1 - (-2), 4 - 1, -1 - 0 >
= < 3, 3, -1 >.
PS = S - P
= < 3 - (-2), 6 - 1, 2 - 0 >
= < 5 , 5, 2 >.
The volume of the parallelepiped can be found using the triple product of these vectors, PS.( PQ * PR )
[tex]= \begin{vmatrix}5 & 5 & 2 \\ 7 & 2 & 5 \\ 3 & 3 & -1\end{vmatrix}[/tex]
We solve this determinant to get the value of the volume of the parallelepiped.
[tex]= \begin{vmatrix}5 & 5 & 2 \\ 7 & 2 & 5 \\ 3 & 3 & -1\end{vmatrix}\\= 5\begin{vmatrix}2 & 5\\ 3 & -1\end{vmatrix} - 5\begin{vmatrix}7 & 5\\ 3 & -1\end{vmatrix} + 2\begin{vmatrix}7 & 2\\ 3 & 3\end{vmatrix}\\[/tex]
= 5(2*(-1)-5*3)-5(7*(-1)-5*3)+2(7*3-2*3)
= 5(-17) - 5(-22) + 2(15)
= -85 + 110 + 30
= 55.
Thus, the volume of the parallelepiped is 55 cubic units.
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