The equation of the line passing through $(1,2,3)$ and perpendicular to the lines…
The equation of the line passing through $(1,2,3)$ and perpendicular to the lines
$x-1=\frac{y+2}{2}=\frac{z+4}{4}$ and $\frac{x-1}{2}=\frac{y-2}{2}=z+3$ is
$\frac{x-1}{6}=\frac{2-y}{7}=\frac{z-3}{2}$
$\frac{x-1}{6}=\frac{y-2}{7}=\frac{z-3}{2}$
$\frac{x-1}{4}=\frac{2-y}{5}=\frac{z-3}{2}$
$x-1=\frac{y-2}{2}=\frac{z-3}{4}$
Solution
The vector perpendicular to both the given lines is given by $\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 2 & 1\end{array}\right|=\hat{i}(-6)-\hat{j}(-7)+\hat{k}(-2)=-6 \hat{i}+7 \hat{j}-2 \hat{k}$
Hence d.r.s. of required line are $6,-7,2$.
Thus eq. of required line is
$\frac{x-1}{6}=\frac{y-2}{-7}=\frac{2-3}{2} \text { i.e. } \frac{x-1}{6}=\frac{2-y}{7}=\frac{z-3}{2}$