The given equation of line is

x−5

3

=

y+4

7

=

6−z

2

i.e in standard form

x−5

3

=

y−(−4)

7

=

z−6

−2

Comparing this equation with standard form

x−x1

a

=

y−y1

b

=

z−z1

c

We get, x1 = 5 , y1 = - 4 , z1 = 6 , a = 3 , b = 7 , c = - 2
Thus, the required line is parallel to the vector 3

^

i

+7

^

j

−2

^

k

and passes through the point (5, -4, 6).
The vector form of the line can be written as

→

r

=

→

a

+λ

→

b

, where λ is a constant
Thus, the required equation is

→

r

= (5

^

i

−4

^

j

+6

^

k

) + λ (3

^

i

+7

^

j

−2

^

k

)