If a line makes angles \(\alpha\), \(\beta\), and \(\gamma\) with the positive directions of the x, y, and z-axis respectively, then \(\cos 2\alpha + \cos 2\beta + \cos 2\gamma\) equals:
If $ | \vec{a} | = 3 $, $ | \vec{b} | = 2 $, then find $ (3\vec{a} - 2\vec{b}) \cdot (3\vec{a} + 2\vec{b}) $.
The value of \( x \) that satisfies the equation:
\[ \begin{vmatrix} x & 1 & 1 \\ 2 & 2 & 0 \\ 1 & 0 & -2 \end{vmatrix} = 6 \]
If \( 2 \) is a solution of the inequality \( \frac{x-a}{a-2x}<-3 \), then \( a \) must lie in the interval:
Let \[ A = \begin{pmatrix} 2 & -1 & 1 \\ -1 & 0 & 2 \\ 1 & -2 & -1 \end{pmatrix} \] and let \( B = \frac{1}{|A|} A \). Then the value of \( |B| \) is equal to:
If $ | \vec{a} | = 3 $, $ | \vec{b} | = 2 $, then find $ (3\vec{a} - 2\vec{b}) \cdot (3\vec{a} + 2\vec{b}) $.