If → a = 2 ^ i + ^ j + 3 ^ k , → b = 3 ^ i + 3 ^ j + ^ k and → c =...

As the given vectors are coplanar, so \left|\begin{cases}2 & 1 & 3 \\ 3 & 3 & 1 \\ {c}_{1} & {c}_{2} & {c}_{3}\end{cases}\right| = 0

$⇒ 8{c}_{1} - 7{c}_{2} - 3{c}_{3} = 0 . . . \left(i\right)$

Also, $\overset{\to}{a} \cdot \overset{\to}{c} = 5 ⇒ 2{c}_{1} + {c}_{2} + 3{c}_{3} = 5 . . . \left(ii\right)$

and $\overset{\to}{b} \cdot \overset{\to}{c} = 0 ⇒ 3{c}_{1} + 3{c}_{2} + {c}_{3} = 0 . . . \left(iii\right)$

Solving the above three equations using Cramer's rule, we get,

∆ = \left|\begin{cases}8 & - 7 & - 3 \\ 2 & 1 & 3 \\ 3 & 3 & 1\end{cases}\right| = - 122

{∆}_{1} = \left|\begin{cases}0 & - 7 & - 3 \\ 5 & 1 & 3 \\ 0 & 3 & 1\end{cases}\right| = - 10

{∆}_{2} = \left|\begin{cases}8 & 0 & - 3 \\ 2 & 5 & 3 \\ 3 & 0 & 1\end{cases}\right| = 85

{∆}_{3} = \left|\begin{cases}8 & - 7 & 0 \\ 2 & 1 & 5 \\ 3 & 3 & 0\end{cases}\right| = - 225

Hence, $122\left({c}_{1} + {c}_{2} + {c}_{3}\right) = 122\left(\dfrac{10}{122} - \dfrac{85}{122} + \dfrac{225}{122}\right) = 150$