Find the value of $\mathrm{\lambda }$, so that the lines $\frac{1-\mathrm{x}}{3}=\frac{7\mathrm{y}-14}{\mathrm{\lambda }}=\frac{\mathrm{z}-3}{2}$ and $\frac{7-7\mathrm{x}}{3\mathrm{\lambda }}=\frac{\mathrm{y}-5}{1}=\frac{6-\mathrm{z}}{5}$ are at right angles. Also, find whether the lines are intersecting or not.

Given lines are $\frac{1-\mathrm{x}}{3}=\frac{7\mathrm{y}-14}{\mathrm{\lambda }}=\frac{\mathrm{z}-3}{2} \mathrm{and} \frac{7-7\mathrm{x}}{3\mathrm{\lambda }}=\frac{\mathrm{y}-5}{1}=\frac{6-\mathrm{z}}{5}$

Converting them into standard form, we have $\frac{\mathrm{x}-1}{-3}=\frac{\mathrm{y}-2}{\left({\displaystyle \raisebox{1ex}{$\mathrm{\lambda }$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}\right)}=\frac{\mathrm{z}-3}{2}$ and $\frac{\mathrm{x}-1}{\left({\displaystyle -\raisebox{1ex}{$3\mathrm{\lambda }$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}\right)}=\frac{\mathrm{y}-5}{1}=\frac{\mathrm{z}-6}{-5}$

Corresponding d.r's are $(\left(-3, \frac{\mathrm{\lambda }}{7}, 2\right)$ and $\left(\frac{-3\mathrm{\lambda }}{7}, 1, -5\right)$

Since the angle between the lines is right angle so,

$$\begin{gathered} \cos {90}^0=\left|\frac{(-3) \left({\displaystyle \frac{-3\mathrm{\lambda }}{7}}\right)+\left({\displaystyle \frac{\mathrm{\lambda }}{7}}\right) \left(1\right)+\left(2\right)(-5)}{\sqrt{{\left(-3\right)}^2+{\left({\displaystyle \frac{\mathrm{\lambda }}{7}}\right)}^2+2^2} \sqrt{{\left({\displaystyle \frac{-3\mathrm{\lambda }}{7}}\right)}^2+1^2+{(-5)}^2}}\right| \\ \Rightarrow 0=\left|\frac{{\displaystyle \frac{9\mathrm{\lambda }}{7}}+{\displaystyle \frac{\mathrm{\lambda }}{7}}-10}{\sqrt{{\displaystyle \frac{{\mathrm{\lambda }}^2}{49}}+13} \sqrt{{\displaystyle \frac{9{\mathrm{\lambda }}^2}{49}}+26}}\right| \end{gathered}$$

$$\begin{gathered} \Rightarrow \left(\frac{10\mathrm{\lambda }}{7}-10\right)=0 \\ \Rightarrow \frac{10\mathrm{\lambda }}{7}=10 \\ \Rightarrow \mathrm{\lambda }=7. \end{gathered}$$

Now, substitute the value of $\mathrm{\lambda }$, $\frac{1-\mathrm{x}}{3}=\frac{\mathrm{y}-2}{7}=\frac{\mathrm{z}-3}{2}=\mathrm{a}, \frac{\mathrm{x}-1}{-3}=\frac{\mathrm{y}-5}{1}=\frac{\mathrm{z}-6}{-5}=\mathrm{b}$  (Assuming)

From first equation, $(\mathrm{x}, \mathrm{y}, \mathrm{z})=(-3\mathrm{a}+1, \mathrm{a}+2, 2\mathrm{a}+3)$ and from second equation $(\mathrm{x}, \mathrm{y}, \mathrm{z})=(-3\mathrm{b}+1, \mathrm{b}+5, -5\mathrm{b}+6)$

Equating the corresponding values of coordinates, we have 

$-3\mathrm{a}+1=-3\mathrm{b}+1, \mathrm{a}+2=\mathrm{b}+5$ and $2\mathrm{a}+3=-5\mathrm{b}+6$ 

$\Rightarrow$ $-3\mathrm{a}+3\mathrm{b}=0, \mathrm{a}-\mathrm{b}=3$ and $2\mathrm{a}+5\mathrm{b}=3$

Solving second and third equations of the above, we get $\mathrm{a}=\frac{18}{7}$ and $\mathrm{b}=\frac{-3}{7}$

Substituting these value of a and b in the first one 

$-3\left(\frac{18}{9}\right)+3\left(\frac{-3}{7}\right)=-9$ 

We can see that the first equation is not satisfied so the lines are not intersecting.
Therefore, the value of $\lambda =7$ and the lines are not intersecting.