Math the following

a) Given equations are consistent if $(\hat{i}+\hat{j})+\lambda (\hat{i}+2\hat{j}-\hat{k})=(\hat{i}+2\hat{j})+\mu (-\hat{i}+\hat{j}+a\hat{k})$

$\Rightarrow \lambda =\frac{{\displaystyle 1}}{3} \text{and }\mu =-\frac{{\displaystyle 1}}{3}\Rightarrow a=1$

$\text{b)}\stackrel{ \to }{a}=\lambda \hat{i}-3\hat{j}-\hat{k}\Rightarrow \overrightarrow{b}=2\lambda \hat{i}+\lambda \hat{j}-\hat{k}$

Angle between $\stackrel{ \to }{a}$ and $\overrightarrow{b}$ is acute.

Therefore, $\overrightarrow{a}\cdot \overrightarrow{b}>0$

$\Rightarrow 2{\lambda }^2-3\lambda +1>0 \text{or }(2\lambda -1)(\lambda -1)>0$

$\text{or }\lambda \in (-\mathrm{\infty },\frac{1}{2})\cup (1,\mathrm{\infty })\text{.}$

Also $\overrightarrow{b}$ makes an obtuse angle with the axes.
Therefore, $\overrightarrow{b}\hat{i}<0\Rightarrow \lambda <0\overrightarrow{b}\cdot \hat{j}<0\Rightarrow \lambda <0....\left(ii\right)$
Combining these two, we get $\lambda =-4,-2$

c) If vectors $2\hat{i}-\hat{j}+\hat{k},\hat{i}+2\hat{j}+(1+a)\hat{k}$ and $3\hat{i}+a\hat{j}+5\hat{k}$ are coplanar, then $\left|\begin{array}{ccc}2& -1& 1\\ 1& 2& 1+a\\ 3& a& 5\end{array}\right|=0$

$\text{or }{a}^2+2a-8=0 \text{or }(a+4)(a-2)=0 \text{or }a=-4,2$

$\text{d)}\Rightarrow \overrightarrow{A}+\lambda \overrightarrow{B}=2(1+\lambda )\hat{i}+(\lambda +{\lambda }^2)\hat{j}+(3+\lambda )\hat{k}$

$\text{Now }(\overrightarrow{A}+\lambda \overrightarrow{B})\perp \overrightarrow{C}\text{. Therefore, }(\overrightarrow{A}+\lambda \overrightarrow{B})\cdot \overrightarrow{C}=0$

$\text{or }6(1+\lambda )+(\lambda +{\lambda }^2)+0=0 \text{or }{\lambda }^2+7\lambda +6=0 \text{or}$

$(\lambda +6)(\lambda +1)=0 \text{or }\lambda =-6,=-1\Rightarrow \left|2\lambda \right|=12,2$