If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+2\hat{k}$,$\vec{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}$ and $\vec{c}$ is a linear combination of $\vec{a}$ and $\vec{b}$,then the value of $x$ is:

  • A
    $1$
  • B
    $-2$
  • C
    $0$
  • D
    $-4$

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If the vectors $\overrightarrow{p}=(a+1) \hat{i}+a \hat{j}+a \hat{k}$,$\overrightarrow{q}=a \hat{i}+(a+1) \hat{j}+a \hat{k}$,and $\overrightarrow{r}=a \hat{i}+a \hat{j}+(a+1) \hat{k}$ $(a \in R)$ are coplanar and $3(\overrightarrow{p} \cdot \overrightarrow{q})^{2}-\lambda|\overrightarrow{r} \times \overrightarrow{q}|^{2}=0$,then the value of $\lambda$ is:

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If $\overrightarrow{a} = 2\hat{i} + \hat{j} + 3\hat{k}$,$\overrightarrow{b} = 3\hat{i} + 3\hat{j} + \hat{k}$ and $\overrightarrow{c} = c_{1}\hat{i} + c_{2}\hat{j} + c_{3}\hat{k}$ are coplanar vectors and $\overrightarrow{a} \cdot \overrightarrow{c} = 5$,$\overrightarrow{b} \perp \overrightarrow{c}$,then $122(c_{1} + c_{2} + c_{3})$ is equal to.......

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