Let the vectors → a = (1 + t) ^ i + (1 - t) ^ j + ^ k , → b = (1 -...

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

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

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

Now, \left[$\begin{cases}\overset{\to}{a} & \overset{\to}{b} & \overset{\to}{c}\end{cases}$\right]$</p><p>$= \left|$\begin{cases}1 + t & 1 - t & 1 \\ 1 - t & 1 + t & 2 \\ t & - t & 1\end{cases}$\right|$</p><p>${C}_{2} \to {C}_{1} + {C}_{2}$</p><p>$= \left|$\begin{cases}1 + t & 2 & 1 \\ 1 - t & 2 & 2 \\ t & 0 & 1\end{cases}$\right|$</p><p>$= 2\left|\begin{cases}1 + t & 1 & 1 \\ 1 - t & 1 & 2 \\ t & 0 & 1\end{cases}\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{cases}\overset{\to}{a} & \overset{\to}{b} & \overset{\to}{c}\end{cases}\right] \ne 0 ⇒ t \ne 0