Line perpendicular to the lines
\(\vec{r}=2 \hat{\imath}+\hat{\jmath}-3 \hat{\mathrm{k}}+\lambda(\hat{\imath}+2 \hat{\jmath}+5 \hat{\mathrm{k}}) \text { and } \vec{r}=3 \hat{\mathrm{\imath}}+3 \hat{\jmath}-7 \hat{\mathrm{k}}+\mu(3 \hat{\imath}-2 \hat{\jmath}+5 \hat{\mathrm{k}})\)
has a vector parallel it is given by
\(\vec{b}=\overrightarrow{b_1} \times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 2 & 5 \\ 3 & -2 & 5\end{array}\right|=20 \hat{\imath}+10 \hat{\jmath}-8 \hat{\mathrm{k}}\)
\(\therefore\) equation of line in vector form is \(\vec{r}=-\hat{\imath}+2 \hat{\jmath}+7 \hat{k}+a(10 \hat{\imath}+5 \hat{\jmath}-4 \hat{\mathrm{k}})\)
And equation of line in cartesian form is \(\frac{x+1}{10}=\frac{y-2}{5}=\frac{z-7}{-4}\)
