Given lines are

1−x

3

=

y−2

2λ

=

z−3

2

and

x−1

3λ

=

y−1

1

=

6−z

7

Let us rewrite the equations of the given lines as follows:

−(x−1)

3

=

y−2

2λ

=

z−3

2

and

x−1

3λ

=

y−1

1

=

−(z−6)

7

That is we have,

x−1

−3

=

y−2

2λ

=

z−3

2

and

x−1

3λ

=

y−1

1

=

z−6

−7

The lines are perpendicular so angle between them is 90
So, cos θ = 0
Here (a1,b1,c1) = (-3,2λ,2) and (a2,b2,c2) = (3λ,1,-7)
For perpendicular lines
a1a2+b1b2+c1c2 = 0
⇒ - 9λ + 2λ - 14 = 0
⇒ - 7λ = 14
⇒ λ =

14

−7

⇒ λ = - 2