Mathematics - Vector Question with Solution | TestHub

$\frac{2-x}{3}=\frac{3y-2}{4\lambda +1}=4-z...(1)$ $\frac{x-2}{(-3)}=\frac{y-\frac{2}{3}}{\left(\frac{4\lambda +1}{3}\right)}=\frac{z-4}{(-1)}$ $\begin{array}{rl}& \frac{x+3}{3\mu }=\frac{1-2y}{6}=\frac{5-z}{7}...(2)\\ & \frac{x+3}{3\mu }=\frac{y-\frac{1}{2}}{(-3)}=\frac{z-5}{(-7)}\\ & \text{ Right angle }\Rightarrow (-3)(3\mu )+\left(\frac{4\lambda +1}{3}\right)(-3)+(-1)(-7)=0\\ & -9\mu -4\lambda -1+7=0\\ & 4\lambda +9\mu =6\end{array}$