If \(\vec{a}, \vec{b}, \vec{c}\) are three vectors mutually perpendicular to each other and \(|\vec{a}|=1,|\vec{b}|=3\) and \(|\vec{c}|=5\), then \([\vec{a}-2 \vec{b} \quad \vec{b}-3 \vec{c} \vec{c}-4 \vec{a}]=\) (1) 0 \((2)-24\) (3) 3600 (4) \(-345\) (5) 15

Vector algebra is a branch of mathematics that deals with vectors, which are quantities that have both magnitude and direction. In this context, we consider operations such as addition, subtraction, and the scalar triple product.

The scalar triple product of three vectors \(\vec{a}, \vec{b}, \vec{c}\), denoted by [\vec{a}, \vec{b}, \vec{c}], is a key concept here. It's defined as the dot product of one of the vectors with the cross product of the other two:

\[( \vec{a} \cdot ( \vec{b} \times \vec{c} ))\]

Understanding how to work with these operations is fundamental to solving the problem correctly.

In this problem, the magnitudes and the mutual perpendicularity of vectors \vec{a}, \vec{b}, \vec{c} play a crucial role.

The determinant is a special number that can be calculated from a square matrix. It's a key concept in linear algebra and is very useful for determining properties of the matrix, such as whether the matrix is invertible.

For three vectors, the scalar triple product gives us the volume of the parallelepiped formed by these vectors. This can be computed using a determinant of a 3x3 matrix:

The matrix is formed by placing each vector's components as rows or columns. For vectors \( \vec{a}, \vec{b}, \vec{c} \),
the determinant of the following matrix gives the scalar triple product:

\[ \text{det}[ \vec{a} \quad \vec{b} \quad \vec{c} ] = \begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix} \]

In our problem, to find \[(\vec{a} - 2 \vec{b}) \cdot ((\vec{b} - 3 \vec{c}) \times ( \vec{c} - 4 \vec{a} )) \], we first form the matrix and compute its determinant.

Mutually perpendicular vectors are vectors that meet at right angles to each other. For three vectors \( \vec{a}, \vec{b}, \vec{c} \) to be mutually perpendicular, their dot products must all be zero:

\[-\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = \vec{b} \cdot \vec{c} = 0\]

Additionally, in vector algebra, these vectors can be used to form an orthogonal basis in three-dimensional space. The fact that \(| \vec{a} | = 1 \), \(| \vec{b} |= 3 \), and \(|\vec{c}| = 5 \) reflects their magnitudes, affecting the computation of the scalar triple product.

The perpendicularity ensures simplifications when computing the cross products and dot products as many terms cancel out. This is the reason why in the given problem, calculations involving the cross product and the determinant become more manageable and lead us to the final solution easily.