We know that, equation of a line passing through $x_1, y_1, z_1$ with direction ratios a, b, c $\frac{x-x_1}{a}$ = $\frac{y-y_1}{b}$ = $\frac{z-z_1}{c}$ So, the required equation of a line passing through ( 1,3, 2) is: $\frac{x+1}{a}$ = $\frac{y-3}{b}$ = $\frac{z+2}{c}$ ... (1) Given that line $\frac{x}{1}$ = $\frac{y}{2}$ = $\frac{z}{3}$ is perpendicular to line (1),so $a_1a_2+b_1b_2+c_1c_2$ = 0 a + 2b + 3c = 0 ... (2) And line $\frac{x+2}{-3}$ = $\frac{y-1}{2}$ = $\frac{z+1}{5}$ is perpendicular to line 1 , so $a_1a_2+b_1b_2+c_1c_2$ = 0 - 3a + 2b + 5c = 0 ... (3) Solving equation 2 and 3 by cross multiplication $\frac{a}{(2)(5)-(2)(3)}$ = $\frac{b}{(-3)(3)-(1)(5)}$ = $\frac{c}{(1)(2)-(-3)(2)}$ ⇒ $\frac{a}{10-6}$ = $\frac{b}{-9-5}$ = $\frac{c}{2+6}$ ⇒ $\frac{a}{4}$ = $\frac{b}{-14}$ = $\frac{c}{8}$ ⇒ $\frac{a}{2}$ = $\frac{b}{-7}$ = $\frac{c}{4}$ = λ (say) ⇒ a = 2λ , b = - 7 λ , c = 4λ Putting the value of a,b, and c in (1) gives $\frac{x+1}{2\lambda}$ = $\frac{y-3}{-7\lambda}$ = $\frac{z+2}{4\lambda}$ ⇒ $\frac{x+1}{2}$ = $\frac{y-3}{-7}$ = $\frac{z+2}{4}$