The equation of the line passing through the point $(3,1,2)$ and perpendicular to the lines…
The equation of the line passing through the point $(3,1,2)$ and perpendicular to the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$ is
$\frac{x+3}{2}=\frac{y+1}{7}=\frac{z+2}{4}$
$\frac{x-3}{-2}=\frac{y-1}{7}=\frac{z-2}{4}$
$\frac{x-3}{2}=\frac{y-1}{-7}=\frac{z-2}{4}$
$\frac{x-3}{2}=\frac{y-1}{5}=\frac{z-2}{4}$
Solution
Required line is perpendicular to the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$.
$\therefore \quad$ Required line is parallel to vector $\bar{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & 2 & 5\end{array}\right|$ $=4 \hat{i}-14 \hat{j}+8 \hat{k}$
$\therefore \quad$ The equation of the required line is $\frac{x-3}{2}=\frac{y-1}{-7}=\frac{z-2}{4}$