Question

Find 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 $\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$.

Answer 1

Equation of the line passing through $(3,1,2)$ is

$\frac{x-3}{a}=\frac{y-1}{b}=\frac{z-2}{c}$ ----- $\left(1\right)$

where $a,b,c$ are the direction ratios of the line

Since, the line is perpendicular to the given lines

So,

$a+2b+3c=0$ ----- $\left(2\right)$

$3a+2b+5c=0$ ---- $\left(3\right)$

Subtracting $\left(3\right)$ from $\left(2\right)$, we get

$-2a-2c=0$

$a=-c$

Putting $a=-c$ in $\left(2\right)$

$-c+2b+3c=0$

$b=-c$

Putting $a=-b$ and $b=-c$ in equation $\left(1\right)$

$\frac{x-3}{-c}=\frac{y-1}{-c}=\frac{z-2}{c}$

$x-3=y-1=-(z-2)$

$x-3=y-1=2-z$

Therefore,

This the required equation of the line.