Let $a, b, c$ be three distinct real numbers,none equal to $1$. If the vectors $a \hat{i}+\hat{j}+\hat{k}$,$\hat{i}+b \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+ c \hat{k}$ are coplanar,then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is equal to

If $a, b,$ and $c$ are coplanar unit vectors,find the value of the scalar triple product $[2a - b, 2b - c, 2c - a]$.

Let the vectors $\vec{a}=(1+t) \hat{i}+(1-t) \hat{j}+\hat{k}$,$\vec{b}=(1-t) \hat{i}+(1+t) \hat{j}+2 \hat{k}$ and $\vec{c}=\hat{i}-t \hat{j}+\hat{k}$,$t \in R$ be such that for $\alpha, \beta, \gamma \in R$,$\alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\vec{0} \Rightarrow \alpha=\beta=\gamma=0$. Then,the set of all values of $t$ is:

Let $a, b$ and $c$ be three non-coplanar vectors and let $p, q$ and $r$ be the vectors defined by $p=\frac{b \times c}{[a b c]}, q=\frac{c \times a}{[a b c]}, r=\frac{a \times b}{[a b c]}$. Then, $(a+b) \cdot p+(b+c) \cdot q+(c+a) \cdot r$ is equal to

If $\hat{i}-3 \hat{j}+\hat{k}$ and $\lambda \hat{i}+3 \hat{j}$ are coplanar with a third vector, let us assume the vectors are $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$, $\vec{b} = \lambda \hat{i}+3 \hat{j}$, and we consider the standard basis vectors or a third vector to define coplanarity. However, if the question implies these two vectors are coplanar with the origin or a specific plane, we evaluate the scalar triple product. Given the standard interpretation of such problems, if $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \lambda \hat{i}+3 \hat{j}$ are coplanar with $\vec{c} = \hat{j}$, then the scalar triple product $[\vec{a} \vec{b} \vec{c}] = 0$. Solving for $\lambda$ where $\vec{a} = (1, -3, 1)$, $\vec{b} = (\lambda, 3, 0)$, and $\vec{c} = (0, 1, 0)$:

If $\overline{a}, \overline{b}$ and $\overline{c}$ are three non-coplanar vectors,then $(\overline{a}+\overline{b}+\overline{c}) \cdot[(\overline{a}+\overline{b}) \times(\overline{a}+\overline{c})]$ equals