Let the vectors $\stackrel{\rightharpoonup }{a}=\left(1+t\right)\hat{i}+\left(1-t\right)\hat{j}+\hat{k},\stackrel{\rightharpoonup }{b}=\left(1-t\right)\hat{i}+\left(1+t\right)\hat{j}+2\hat{k} \mathrm{and} \stackrel{\rightharpoonup }{c}=t\hat{i}-t\hat{j}+\hat{k}, t\in R$

be such that for $\alpha , \beta , \gamma \in R,\alpha \overrightarrow{a}+\beta \overrightarrow{b}+\gamma \overrightarrow{c}=\overrightarrow{O ,}\Rightarrow \alpha =\beta =\gamma =0.$ Then the set of all values of t is

By its given condition

$$\begin{gathered} : \stackrel{\rightharpoonup }{a}, \stackrel{\rightharpoonup }{b}, \stackrel{\rightharpoonup }{c} are linerarly independent vectors \\ \Rightarrow \left[\overline{a} \overline{b} \overline{c}\right]\ne 0 ....\left(i\right) \\ Now, \left[\overline{a} \overline{b} \overline{c}\right] \\ =\left|\begin{array}{ccc}1+t& 1-t& 1\\ 1-t& 1+t& 2\\ t& -t& 1\end{array}\right| \\ {C}_2\to C_1+C_2 \\ \left|\begin{array}{ccc}1+t& 2& 1\\ 1-t& 2& 2\\ t& 0& 1\end{array}\right| \\ =2\left|\begin{array}{ccc}1+t& 1& 1\\ 1-t& 1& 2\\ t& 0& 1\end{array}\right| \\ =2\left[\left(1+t\right)-\left(1-t\right)+t\right] \\ =2\left[3t\right]=6t. \\ \sin ce \left[\overline{a} \overline{b} \overline{c}\right]\ne 0\Rightarrow t\ne 0 \end{gathered}$$