The value of $[a - b, b - c, c - a]$,where $|a| = 1, |b| = 5$ and $|c| = 3$,is

If $\hat{i}-3 \hat{j}+\hat{k}$ and $\lambda \hat{i}+3 \hat{j}$ are coplanar with a third vector,assuming the question implies the vectors are linearly dependent or part of a coplanar set,find $\lambda$. Given the standard form of such problems,if we consider the vectors $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \lambda \hat{i}+3 \hat{j}$ to be coplanar with a reference vector,let us assume the third vector is $\hat{k}$. For these to be coplanar,the scalar triple product must be zero: $\left|\begin{array}{ccc} 1 & -3 & 1 \\ \lambda & 3 & 0 \\ 0 & 0 & 1 \end{array}\right| = 0$. Solving this,$\lambda$ is equal to:

If $\left[ {\vec a \,\vec b \,\vec c } \right] = 4$,then $\left[ {\vec a \times \vec b, \vec b \times \vec c, \vec c \times \vec a } \right] = \dots$

Given vectors $a, b, c$ such that $a \cdot (b \times c) = \lambda \neq 0$,the value of $\frac{(b \times c) \cdot (a + b + c)}{\lambda}$ is

Let $\vec{b} = -\hat{i} + 4\hat{j} + 6\hat{k}$ and $\vec{c} = 2\hat{i} - 7\hat{j} - 10\hat{k}$. If $\vec{a}$ is a unit vector and the scalar triple product $[\vec{a} \ \vec{b} \ \vec{c}]$ has the greatest value,then $\vec{a}$ is equal to:

If $a$ is perpendicular to $b$ and $c$,$|a| = 2$,$|b| = 3$,$|c| = 4$ and the angle between $b$ and $c$ is $\frac{2\pi}{3}$,then $[a \; b \; c]$ is equal to (in $\sqrt{3}$):