The equation of a line passing through the point $(2,1,3)$ and perpendicular to the lines…
The equation of a line passing through the point $(2,1,3)$ 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
$\frac{x-2}{-2}=\frac{y-1}{7}=\frac{z-3}{4}$
$\frac{x-2}{2}=\frac{1-y}{7}=\frac{z-3}{4}$
$\frac{x-2}{2}=\frac{y-1}{4}=\frac{z-3}{7}$
$\frac{x-2}{2}=\frac{1-y}{4}=\frac{z-3}{7}$
Solution
d.r's of perpendicular to both the given lines can be obtained by
$\begin{aligned}
& \frac{a}{2 \times 5-2 \times 3}=\frac{b}{-3 \times 3-1 \times 5}=\frac{c}{1 \times 2-(-3) \times 2} \\
& \Rightarrow \frac{a}{4}=\frac{b}{-14}=\frac{c}{8} \\
& \Rightarrow \text { d.r's are } < 2,-7,4>
\end{aligned}$
now, required equation of the line $\frac{x-2}{2}=\frac{y-1}{-7}=\frac{z-3}{4}$
$\Rightarrow \frac{x-2}{2}=\frac{1-y}{7}=\frac{z-3}{4}$