Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\vec{a} \times \vec{b}=2(\vec{a} \times \vec{c})$. If $|\vec{a}|=1,|\vec{b}|=4,|\vec{c}|=2$, and the angle between $\vec{b}$ and $\vec{c}$ is $60^{\circ}$, then $|\vec{a} \cdot \vec{c}|$ is equal to

Let $\vec{a}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \vec{b}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}, \vec{c}=\lambda \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\vec{v}=\vec{a} \times \vec{b}$. If $\vec{v} \cdot \vec{c}=11$ and the length of the projection of $\vec{b}$ on $\vec{c}$ is $p$, then $9 p^2$ is equal to

Let $\overrightarrow{\mathrm{a}}=-\hat{i}+\hat{j}+2 \hat{k}, \overrightarrow{\mathrm{~b}}=\hat{i}-\hat{j}-3 \hat{k}, \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}$. Then $(\vec{a}-\vec{b}) \cdot \vec{d}$ is equal to :

Let $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\lambda \hat{j}+2 \hat{k}, \lambda \in \boldsymbol{Z}$ be two vectors. Let $\vec{c}=\vec{a} \times \vec{b}$ and $\vec{d}$ be a vector of magnitude 2 in $y z$-plane. If $|\vec{c}|=\sqrt{53}$, then the maximum possible value of $(\vec{c} \cdot \vec{d})^2$ is equal to :