If \[\overrightarrow a = - 3\widehat i + 7\widehat j + 5\widehat k\]\[\overrightarrow b = - 3\widehat i + 7\widehat j - 3\widehat k\]\[\overrightarrow c = 7\widehat i - 5\widehat j - 3\widehat k\] are the three coterminous edges of a parallelepiped, then its volume is
A. \[108\]
B. \[210\]
C. \[272\]
D. \[308\]

Hint: In our case, we have to utilize the fact that V=AH determines the volume of a parallelepiped with a base area of ‘A’ and a height of ‘H’. Consequently, ascertain the parallelepiped's volume. We have to use the volume of a parallelepiped with coterminous edges as \[\vec a,\vec b,\vec c\] given by \[V = \left[ {\vec a,\vec b,\vec c} \right]\] as an alternative; we have \[\left[ {\vec a,\vec b,\vec c} \right]\] being the scalar triple product of the vectors a, b, and c.



Formula Used:Volume of parallelepiped with coterminous can be calculated using
\[V = \left[ {\vec a,\vec b,\vec c} \right]\]



Complete step by step solution:Here, we have been given the three coterminous edges of a parallelepiped as
\[\overrightarrow a = - 3\widehat i + 7\widehat j + 5\widehat k\]
\[\overrightarrow b = - 3\widehat i + 7\widehat j - 3\widehat k\]
\[\overrightarrow c = 7\widehat i - 5\widehat j - 3\widehat k\]
And we are asked to determine the volume of the parallelepiped.
We have been already known that the volume of parallelepiped with coterminous edges is given by
\[V = [\overrightarrow a ,\overrightarrow b ,\overrightarrow c ]\]
We also know that,
If
\[\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k,\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k,\vec c = {c_1}\hat i + {c_2}\hat j + {c_3}\hat k\]
Then,
\[[\vec a,\vec b,\vec c] = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right|\]
Therefore, we have
Volume \[ = \left| {\begin{array}{*{20}{c}}{ - 3}&7&5\\{ - 3}&7&{ - 3}\\7&{ - 5}&{ - 3}\end{array}} \right|\]
Now, we have to use the expansion formula to expand the determinant, we have
\[ = \left( { - 3} \right)\det \left( {\begin{array}{*{20}{c}}7&{ - 3}\\{ - 5}&{ - 3}\end{array}} \right) - 7 \cdot \det \left( {\begin{array}{*{20}{c}}{ - 3}&{ - 3}\\7&{ - 3}\end{array}} \right) + 5 \cdot \det \left( {\begin{array}{*{20}{c}}{ - 3}&7\\7&{ - 5}\end{array}} \right)\]
Now, let us calculate the minors, we obtain
\[ = \left( { - 3} \right)\left( { - 21 - (15)} \right) - (7)(9 + 21) + 5\left( {15 - 49} \right)\]
On simplifying the terms inside the parentheses in the above expression, we get
\[ = \left( { - 3} \right)\left( { - 36} \right) - 7(30) + 5\left( { - 34} \right)\]
Now, let us multiply each term outside with the term inside the parentheses, we get
\[ = 108 - 210 - 170\]
On adding/subtracting the above expression, we get
\[ = - 272\]
As we already know, that volume will always be positive.
Therefore, the volume of a parallelepiped is \[272\]cubic units.



Option ‘C’ is correct



Note:
Students should keep in mind that the three vectors are thought to be coplanar if it equals zero. The cross product is distributive over addition and anti-commutative. So, one should be very cautious while solving these types of problems because it includes formulas and scalar concepts to be remembered in order to solve this problem to get the desired answer.