Two adjacent sides of a parallelogram $ABCD$ are given by $\overrightarrow{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\overrightarrow{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side $AD$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that $AD$ becomes $AD'$. If $AD'$ makes a right angle with the side $AB$,then the cosine of the angle $\alpha$ is given by

Let $\vec{u}, \vec{v}$ and $\vec{w}$ be vectors such that $|\vec{u}+\vec{v}+\vec{w}|=0$. If $|\vec{u}|=3$,$|\vec{v}|=4$ and $|\vec{w}|=5$,then the value of $|\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}|$ is

Let $A=(3,4,0), B=(4,4,4), C=(-6,2,3)$ and $D=(1,1,2)$. If $\theta$ is the acute angle between the lines $AB$ and $CD$, then $\cos \theta=$

Find the distance between the lines $l_{1}$ and $l_{2}$ given by $\vec{r}=\hat{i}+2 \hat{j}-4 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$ and $\vec{r}=3 \hat{i}+3 \hat{j}-5 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+6 \hat{k})$.

Let $ABC$ be a triangle such that $\overrightarrow{BC} = \overrightarrow{a}$,$\overrightarrow{CA} = \overrightarrow{b}$,$\overrightarrow{AB} = \overrightarrow{c}$,$|\overrightarrow{a}| = 6\sqrt{2}$,$|\overrightarrow{b}| = 2\sqrt{3}$,and $\overrightarrow{b} \cdot \overrightarrow{c} = 12$. Consider the statements:
$(S1): |(\overrightarrow{a} \times \overrightarrow{b}) + (\overrightarrow{c} \times \overrightarrow{b})| - |\overrightarrow{c}| = 6(2\sqrt{2} - 1)$
$(S2): \angle ABC = \cos^{-1}\left(\sqrt{\frac{2}{3}}\right)$.
Which of the following is true?

If $\vec{a} = 2\hat{i} - \hat{j} - 2\hat{k}$ and $\vec{b} = 6\hat{i} + 2\hat{j} - 3\hat{k}$ are two vectors,and we consider a vector $\vec{c} = \vec{a} + t\vec{b}$,find the value of $t$ such that the magnitude $|\vec{c}|$ is minimum.