Maths Determinants

$\Delta =0$

$\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 5\hfill & \hfill \alpha \hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \end{array}\right|=0$

15 - 2a + a - 6 – 1 = 0

a = 8

For a = 8, equations are

x + y + 3 = 6

2x + 5y + 8z = b

x + 2y + 3z = 14

$\left(2,5,8\right)=\mathcal{l}\left(1,1,1\right)+m\left(1,2,3\right)$

$\begin{array}{cc}\hfill 2=\mathcal{l}+m\hfill & \hfill 5=\mathcal{l}+2m\hfill \end{array}]\to 3=m,\mathcal{l}=-1$              

8 =  $\mathcal{l}+3m$

$\beta =6\mathcal{l}+14m$

= -6 + 42 = 36

a + b = 8 + 36 = 44

System of equation can be written as

$\left(\begin{array}{ccc}\hfill 2\hfill & \hfill -3\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill {\lambda }^2-\left|\lambda \right|\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}9\\ -18\\ 16\end{array}\right)$

for no solution

$\frac{9}{10}\left({\lambda }^2-\left|\lambda \right|+3\right)-7=0$

$\Rightarrow 9{\lambda }^2-9\left|\lambda \right|-43=0$

$\Rightarrow \left|\lambda \right|=\frac{9+\sqrt[]{81+36*43}}{18}>0$

$\therefore$ only two possible value of $\lambda$ are there.

 $\overrightarrow{a}*\left(\overrightarrow{b}*\overrightarrow{c}\right)=3\overrightarrow{b}-\overrightarrow{c}=\overrightarrow{u}$

$\overrightarrow{b}*\left(\overrightarrow{c}*\overrightarrow{a}\right)=\overrightarrow{c}-2\overrightarrow{a}=\overrightarrow{v}$

$\overrightarrow{c}*\left(\overrightarrow{b}*\overrightarrow{a}\right)=3\overrightarrow{b}-2\overrightarrow{a}=\overrightarrow{w}$

$\therefore \overrightarrow{u}+\overrightarrow{v}=\overrightarrow{w}$

so vectors 

$\overrightarrow{u},\overrightarrow{v}and\overrightarrow{w}$

are coplanar, hence their Scalar triple product will be zero.

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