Let the equation of plane be ax + by + cz + d = 0 …. (1)
Since the plane passes through the point A (0, 0, 0) and B(3, -1, 2), we have a × 0 + b × 0 + c × 0 + d = 0
⇒ d = 0 …. (2)
Similarly for point B (3, -1 , 2), a x 3 + b x (-1) + c x 2 + d = 0
3a – b + 2c = 0 (Using, d = 0) …(3)
Given equation of the line is

x−4

1

=

y+3

−4

=

z−1

7

We can also write the above equation as

x−4

1

=

y−(−3)

−4

=

x−(−1)

7

The required plane is parallel to the above line.
Therefore, a × 1 + b × (-4) + c × 7= 0
⇒ a – 4b + 7c = 0 … (4)
Cross multiplying equations (3) and (4), we obtain:

a

(−1)×7−(−4)×2

=

b

2×1−3×7

=

c

3×(−4)−1×(−1)

⇒

a

−7+8

=

b

2−21

=

c

−12+1

⇒

a

1

=

b

−19

=

c

−11

= k
⇒ a = k , b = - 19k , c = - 11k
Substituting the values of a, b and c in equation (1), we obtain the equation of plane as:
kx - 19ky - 11kz + d = 0
⇒ k (x - 19y - 11z) = 0 (From equation(2))
⇒ x - 19y - 11z = 0
So, the equation of the required plane is x – 19y - 11z = 0