An algebra problem by Ankit Nigam

$\left| \overrightarrow { c } \right| =\sqrt { 3 }$

$1+{ \alpha }^{ 2 }+{ \beta }^{ 2 }=3$

$\therefore { \alpha }^{ 2 }+{ \beta }^{ 2 }=2$

$since\quad \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } are\quad linearly\quad dependent$

$\Rightarrow \overrightarrow { c } =x\overrightarrow { a } +y\overrightarrow { b }$

$\Rightarrow \hat { i } +\alpha \hat { j } +\beta \hat {k} =x\left( \overrightarrow { i } +\overrightarrow { j } +\overrightarrow { k } \right) +\left( 4\overrightarrow { i } +3\overrightarrow { j } +4\overrightarrow { k } \right)$

$comparing\quad equations\quad we\quad get\quad \\ x+4y=1\\ x+3y=\alpha \\ x+4y=\beta$

$\therefore \beta =1\quad and\quad from\quad { \alpha }^{ 2 }+{ \beta }^{ 2 }=2\\ \quad we\quad get\quad \alpha =\pm 1\quad$