Let the vectors a → = 1 + t i ^ + 1 - t j ^ + k ^ , b → = 1 - t i ^ + 1 + t j ^ + 2 k ^ and c…

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

By its given condition $:\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$are linearly independent vectors or they are non-coplanar

We know that when vectors are non-coplanar then, $\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]\ne 0 ...\left(1\right)$

Now, $\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]$

$=\left|\begin{array}{ccc}1+t& 1-t& 1\\ 1-t& 1+t& 2\\ t& -t& 1\end{array}\right|$

$C_2\to C_1+C_2$

$=\left|\begin{array}{ccc}1+t& 2& 1\\ 1-t& 2& 2\\ t& 0& 1\end{array}\right|$

$=2\left|\begin{array}{ccc}1+t& 1& 1\\ 1-t& 1& 2\\ t& 0& 1\end{array}\right|$

$=2\left[\left(1+t\right)-\left(1-t\right)+t\right]$

$=2\left[3t\right]=6t$

Now from equation $\left(1\right)$ we get, $\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]\ne 0\Rightarrow t\ne 0$