In a $\triangle ABC$,let $G$ denote its centroid and let $M, N$ be points in the interiors of the segments $AB, AC$,respectively,such that $M, G, N$ are collinear. If $r$ denotes the ratio of the area of $\triangle AMN$ to the area of $\triangle ABC$,then

Let $\lambda \in \mathbb{Z}$,$\vec{a} = \lambda \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}$. Let $\vec{c}$ be a vector such that $(\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = \vec{0}$,$\vec{a} \cdot \vec{c} = -17$ and $\vec{b} \cdot \vec{c} = -20$. Then $|\vec{c} \times (\lambda \hat{i} + \hat{j} + \hat{k})|^2$ is equal to

If $|a|=3, |b|=4$ and the angle between $a$ and $b$ is $120^{\circ}$,then $|4a+3b|$ is equal to

If $\overrightarrow{a}=\hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}-3 \hat{j}+\hat{k}$ and the orthogonal projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is $\frac{4}{3}(\hat{i}-\hat{j}-\hat{k})$,then $\lambda$ is equal to

If $\theta$ is the angle between any two vectors $\vec{a}$ and $\vec{b},$ then $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$ when $\theta$ is equal to

Let $\vec{a} = \lambda \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} + 4\hat{j} + 4\hat{k}$, and $\vec{c} = \hat{i} + \mu \hat{j} + \hat{k}$. If $\vec{a}$ is parallel to $\vec{b}$ and $\vec{b}$ is perpendicular to $\vec{c}$, then find the value of $\lambda - \mu$.