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

x−x1

a

=

y−y1

b

=

z−z1

c

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

x+1

a

=

y−3

b

=

z+2

c

... (1)
Given that line

x

1

=

y

2

=

z

3

is perpendicular to line (1),so
a1a2+b1b2+c1c2 = 0
a + 2b + 3c = 0 ... (2)
And line

x+2

−3

=

y−1

2

=

z+1

5

is perpendicular to line 1 , so
a1a2+b1b2+c1c2 = 0
- 3a + 2b + 5c = 0 ... (3)
Solving equation 2 and 3 by cross multiplication

a

(2)(5)−(2)(3)

=

b

(−3)(3)−(1)(5)

=

c

(1)(2)−(−3)(2)

⇒

a

10−6

=

b

−9−5

=

c

2+6

⇒

a

4

=

b

−14

=

c

8

⇒

a

2

=

b

−7

=

c

4

= λ (say)
⇒ a = 2λ , b = - 7 λ , c = 4λ
Putting the value of a,b, and c in (1) gives

x+1

2λ

=

y−3

−7λ

=

z+2

4λ

⇒

x+1

2

=

y−3

−7

=

z+2

4