If $\bar{a}=2 \hat{i}-\hat{j}+\hat{k}, \bar{b}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\bar{c}=3 \hat{i}+\lambda…
If $\bar{a}=2 \hat{i}-\hat{j}+\hat{k}, \bar{b}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\bar{c}=3 \hat{i}+\lambda \hat{j}+5 \hat{k}$ are coplanar, then $\lambda$ is the root of the equation
$x^2+3 x=6$
$x^2+2 x=4$
$x^2+3 x=4$
$x^2+2 x=6$
Solution
Since given vectors are coplanar, we write
$\begin{aligned}
& \left|\begin{array}{ccc}
2 & -1 & 1 \\
1 & 2 & -3 \\
3 & \lambda & 5
\end{array}\right|=0 \\
& \therefore 2(10+3 \lambda)+1(5+9)+1(\lambda-6)=0 \\
& \therefore 20+6 \lambda+14+\lambda-6=0 \Rightarrow 7 \lambda+28=0 \Rightarrow \lambda=-4
\end{aligned}$
Which is one of the root of equation in option $C$