If $\overline{a}=3 \hat{i}-5 \hat{j}$ and $\overline{b}=6 \hat{i}-3 \hat{j}$ are two vectors and $\overline{c}$ is a vector such that $\overline{c}=\overline{a} \times \overline{b}$,then the ratio $a: b: c$ is:

$A$ force $\overrightarrow{F} = 2i + j - k$ acts at a point $A$,whose position vector is $2i - j$. The moment of $\overrightarrow{F}$ about the origin is:

If $\bar{a} = \bar{i} - 2\bar{j} - 2\bar{k}$ and $\bar{b} = 2\bar{i} + \bar{j} + 2\bar{k}$ are two vectors, then $(\bar{a} + 2\bar{b}) \times (3\bar{a} - \bar{b}) = $

If $a+2b+3c=0$ and $(a \times b)+(b \times c)+(c \times a)=\lambda(b \times c)$,then the value of $\lambda$ is equal to

If either $\vec{a}=\overrightarrow{0}$ or $\vec{b}=\overrightarrow{0},$ then $\vec{a} \times \vec{b}=\overrightarrow{0} .$ Is the converse true? Justify your answer with an example.

Let $\lambda \in R$,$\vec{a} = \lambda \hat{i} + 2 \hat{j} - 3 \hat{k}$,and $\vec{b} = \hat{i} - \lambda \hat{j} + 2 \hat{k}$. If $((\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b})) \times (\vec{a} - \vec{b}) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$,then $|\lambda(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})|^2$ is equal to