Find the equation of the line passing through the point  (-1,3,-2)  and

perpendicular to the lines   $\frac{\mathrm{x}}{1} = \frac{\mathrm{y}}{2} = \frac{\mathrm{z}}{3} \mathrm{and} \frac{\mathrm{x} + 2}{- 3} = \frac{\mathrm{y} - 1}{2} = \frac{\mathrm{z} + 1}{5}.$

We know that, equation of a line passing through  x₁,  y₁,  z₁  with direction ratios  a,  b,  c

$\mathrm{Is} \mathrm{given} \mathrm{by} \frac{\mathrm{x} - {\mathrm{x}}_1}{\mathrm{a}} = \frac{\mathrm{y} - {\mathrm{y}}_1}{\mathrm{b}} = \frac{\mathrm{z} - {\mathrm{z}}_1}{\mathrm{c}}$

So, the required equation of a line passing through ( - 1,  3,  - 2 ) is:

$$\begin{gathered} \frac{\mathrm{x} + 1}{\mathrm{a}} = \frac{\mathrm{y} - 3}{\mathrm{b}} = \frac{\mathrm{z} + 2}{\mathrm{c}} ..............( \mathrm{i} ) \\ \mathrm{Given} \mathrm{that} \mathrm{the} \mathrm{line} \frac{\mathrm{x}}{1} = \frac{\mathrm{y}}{2} = \frac{\mathrm{z}}{3} \mathrm{is} \mathrm{perpendicular} \mathrm{to} \mathrm{line} ( \mathrm{i} ), \mathrm{so} \\ {\mathrm{a}}_1 {\mathrm{a}}_2 + {\mathrm{b}}_1 {\mathrm{b}}_{2 } + {\mathrm{c}}_1 {\mathrm{c}}_2 = 0 \\ \mathrm{a} \mathrm{x} 1 + \mathrm{b} \mathrm{x} 2 + \mathrm{c} \mathrm{x} 3 = 0 \\ \mathrm{a} + 2 \mathrm{b} + 3 \mathrm{c} = 0 .............( \mathrm{ii} ) \\ \mathrm{And} \mathrm{line} \frac{ \mathrm{x} + 2}{- 3} = \frac{\mathrm{y} - 1}{2} = \frac{\mathrm{z} + 1}{5} \mathrm{is} \mathrm{perpendicular} \mathrm{to} \mathrm{line} ( \mathrm{i} ), \mathrm{so} \\ {\mathrm{a}}_1 {\mathrm{a}}_2 + {\mathrm{b}}_1 {\mathrm{b}}_{2 } + {\mathrm{c}}_1 {\mathrm{c}}_2 = 0 \\ \mathrm{a} \mathrm{x} ( - 3 ) + \mathrm{b} \mathrm{x} 2 + \mathrm{c} \mathrm{x} 5 = 0 \\ - 3 \mathrm{a} + 2 \mathrm{b} + 5 \mathrm{c} = 0 ............( \mathrm{iii} ) \end{gathered}$$

Solving equation  ( ii )  and  ( iii )  by cross multiplication,

$$\begin{gathered} \frac{\mathrm{a}}{2 \mathrm{x} 5 - 2 \mathrm{x} 3} = \frac{\mathrm{b}}{( - 3 ) \mathrm{x} 3 - 1 \mathrm{x} 5} = \frac{\mathrm{c}}{1 \mathrm{x} 2 - ( - 3 ) \mathrm{x} 2} \\ \Rightarrow \frac{\mathrm{a}}{10 - 6} = \frac{\mathrm{b}}{- 9 - 5} = \frac{\mathrm{c}}{2 + 6} \\ \Rightarrow \frac{\mathrm{a}}{4} = \frac{\mathrm{b}}{- 14} = \frac{\mathrm{c}}{8} \\ \Rightarrow \frac{{\displaystyle \mathrm{a}}}{{\displaystyle 2}} = \frac{{\displaystyle \mathrm{b}}}{{\displaystyle - 7}} = \frac{{\displaystyle \mathrm{c}}}{{\displaystyle 4}} = \mathrm{\lambda } ( \mathrm{say} ) \\ \Rightarrow \mathrm{a} = 2 \mathrm{\lambda }, \mathrm{b} = - 7 \mathrm{\lambda }, \mathrm{c} = 4 \mathrm{\lambda } \end{gathered}$$

Putting the value of  a,  b  and  c  in  ( i ) gives

$$\begin{gathered} \frac{\mathrm{x} + 1}{2 \mathrm{\lambda } } = \frac{\mathrm{y} - 3}{- 7 \mathrm{\lambda } } = \frac{\mathrm{z} + 2}{4 \mathrm{\lambda }} \\ \Rightarrow \frac{{\displaystyle \mathrm{x} + 1}}{{\displaystyle 2 }} = \frac{{\displaystyle \mathrm{y} - 3}}{{\displaystyle - 7 }} = \frac{{\displaystyle \mathrm{z} + 2}}{{\displaystyle 4 }} \end{gathered}$$