Let → a = 2 ^ i + ^ j - ^ k and → b = ^ i + 2 ^ j + ^ k be two...

$\begin{cases}∵ \overset{\to}{a} + \overset{\to}{b} = 3i + 3j & ∵ \overset{\to}{a} = 2i + j - k, & \left|\overset{\to}{a}\right| = \sqrt{6}\end{cases}$
$\begin{cases}⇒ \left|\overset{\to}{a} + \overset{\to}{b}\right| = 3\sqrt{2}, & \overset{\to}{b} = i + 2j + k, & \left|\overset{\to}{b}\right| = \sqrt{6}\end{cases}$
$\overset{\to}{a} . \overset{\to}{b} = 2 + 2 - 1 = 3$
Now,
as given projection of $\overset{\to}{a} + \overset{\to}{b}$ on $\overset{\to}{c} = 3\sqrt{2}$
$\dfrac{\left(\overset{\to}{a} + \overset{\to}{b}\right)}{\left|\overset{\to}{a} + \overset{\to}{b}\right|} . \overset{\to}{c} = 3\sqrt{2}$
\left(\overset{\to}{a} + \overset{\to}{b}\right) . \left(\alpha\overset{\to}{a} + \beta\overset{\to}{b}\right) = 18$<br />$\alpha{\left|\overset{\to}{a}\right|}^{2} + \alpha\left(\overset{\to}{a} . \overset{\to}{b}\right) + \beta{\left|\overset{\to}{b}\right|}^{2} + \beta\left(\overset{\to}{a} . \overset{\to}{b}\right) = 18$<br />as given$6\alpha + 3\alpha + 6\beta + 3\beta = 18$<br />$\begin{cases}\overset{\to}{c} = \alpha\overset{\to}{a} + \beta\overset{\to}{b}, & \alpha + \beta = 2\end{cases}&hellip;(i)<br />Since\overset{\to}{a}, \overset{\to}{b}$and$\overset{\to}{c}$are co-planar hence \left(\overset{\to}{a} \times \overset{\to}{b}\right) . \overset{\to}{c} = 0$
Minimum value of \left(\overset{\to}{c} - \left(\overset{\to}{a} \times \overset{\to}{b}\right)\right) . \overset{\to}{c} =$Minimum of$\overset{\to}{c} . \overset{\to}{c}$<br />= Minimum of${\left|\overset{\to}{c}\right|}^{2}$<br />\left(∵ \left(\overset{\to}{a} \times \overset{\to}{b}\right) . \overset{\to}{c} = 0\right)$
now $\overset{\to}{c} = \alpha\overset{\to}{a} + \beta\overset{\to}{b}$ and $\alpha + \beta = 2$
$\overset{\to}{c} = \alpha\left(2i + j - k\right) + \left(2 - \alpha\right)\left(i + 2j + k\right) =$
$\overset{\to}{c} = \left(2 + \alpha\right)i + \left(4 - \alpha\right)j + k\left(2 - 2\alpha\right)$
${\left|\overset{\to}{c}\right|}^{2} = 6\left({\alpha}^{2} - 2\alpha + 4\right)$
$\begin{cases}Minimum{\left|\overset{\to}{c}\right|}^{2} = 18 & \left(∴ Minimumofa{n}^{2} + bn + c = - \dfrac{D}{4a} = - \dfrac{{b}^{2} - 4ac}{4a}\right)\end{cases}$