Find the value of k so that the lines $\mathrm{x}=-\mathrm{y}=\mathrm{kz}\text{ and }\mathrm{x}-2=2\mathrm{y}+1=-\mathrm{z}+1$ are perpendicular to each
other.

Given that,
$\begin{array}{l}\overrightarrow{{\mathrm{L}}_1}=\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{\frac{1}{\mathrm{k}}}\\ \overrightarrow{{\mathrm{L}}_2}=\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}+\frac{1}{2}}{\frac{1}{2}}=\frac{\mathrm{z}-1}{-1}\end{array}$
Now, to prove The lines are perpendicular to each other
Direction ratio ${\mathrm{L}}_1=\left(1,-1,\frac{1}{\mathrm{k}}\right)$
Direction ratio ${\mathrm{L}}_2=\left(1,\frac{1}{2},-1\right)$
Now, we know that two lines are perpendicular when dot product of their direction cosines is zero
$\therefore {\mathrm{L}}_1\cdot {\mathrm{L}}_2=0$
$\begin{array}{l}\therefore \left(1,-1,\frac{1}{\mathrm{k}}\right)\cdot \left(1,\frac{1}{2},-1\right)=0\\ \therefore 1\times 1+(-1)\times \frac{1}{2}+\frac{1}{\mathrm{k}}\times (-1)=0\\ \therefore 1-\frac{1}{2}-\frac{1}{\mathrm{k}}=0\\ \therefore \frac{1}{2}=\frac{1}{\mathrm{k}}\\ \therefore \mathrm{k}=2\end{array}$
Therefore, the value of k is 2.