The sum of all values of α , for which the points whose position...

Let the given vectors be \overset{\to}{A} = \overset{^}{i} - 2\overset{^}{j} + 3\overset{^}{k}, \overset{\to}{B} = 2\overset{^}{i} - 3\overset{^}{j} + 4\overset{^}{k}, \overset{\to}{C} = \left(\alpha + 1\right)\overset{^}{i} + 2\overset{^}{k}and\overset{\to}{D} = 9\overset{^}{i} + \left(\alpha - 8\right)\overset{^}{j} + 6\overset{^}{k}

We know that if the vectors are coplanar then \left[$\begin{cases}\overset{\to}{AB} & \overset{\to}{AC} & \overset{\to}{AD}\end{cases}$\right] = 0$</p><p>$⇒ \overset{\to}{AB} = \left(2\overset{^}{i} - 3\overset{^}{j} + 4\overset{^}{k}\right) - \left(\overset{^}{i} - 2\overset{^}{j} + 3\overset{^}{k}\right) = \overset{^}{i} - \overset{^}{j} + \overset{^}{k}$</p><p>$⇒ \overset{\to}{AC} = \left(\left(\alpha + 1\right)\overset{^}{i} + 2\overset{^}{k}\right) - \left(\overset{^}{i} - 2\overset{^}{j} + 3\overset{^}{k}\right) = \alpha\overset{^}{i} + 2\overset{^}{j} - \overset{^}{k}$</p><p>$⇒ \overset{\to}{AD} = \left(9\overset{^}{i} + \left(\alpha - 8\right)\overset{^}{j} + 6\overset{^}{k}\right) - \left(\overset{^}{i} - 2\overset{^}{j} + 3\overset{^}{k}\right) = 8\overset{^}{i} + \left(\alpha - 6\right)\overset{^}{j} + 3k$</p><p>Now,</p><p>$⇒ \left|$\begin{cases}1 & - 1 & 1 \\ \alpha & 2 & - 1 \\ 8 & \alpha - 6 & 3\end{cases}$\right| = 0$</p><p>$⇒ 1\left(6 + \alpha - 6\right) + \left(3\alpha + 8\right) + \left({\alpha}^{2} - 6\alpha - 16\right) = 0$</p><p>On Simplifying we get,</p><p>$⇒ {\alpha}^{2} - 2\alpha - 8 = 0$</p><p>Therefore, sum of the roots is$- \left(- 2\right) = 2.