It is known that the equation of the line passing through the points (x1,y1,z1) and (x2,y2,z2) is
‌

x−x1

x2−x1

=‌

y−y1

y2−y1

=‌

z−z1

z2−z1

So here we have,
x1=3,y1=−4,z1=−5‌ and ‌x2=2,y2=−3,z2=1
The line passing through the points, (3,−4,−5) and (2,−3,1) is given by,
‌

x−3

2−3

=‌

y+4

−3+4

=‌

z+5

1+5

Let ‌

x−3

−1

=‌

y+4

1

=‌

z+5

6

=k
Now,
⇒‌

x−3

−1

=k
∴x=−k+3
⇒‌

y+4

1

=k
∴y=k−4
⇒‌

z+5

6

=k
∴z=6k−5
The co-ordinates of the point of intersection of the given line and plane is, (−k+3,k−4,6k−5)
If is lies on the plane 2x+y+z=7, then,
⇒2(−k+3)+k−4+6k−5=7
⇒5k=10
∴k=2
Substituting k = 2 in ( 1 ) we get,
The co-ordinates of the point of intersection of the given line and plane are
(−k+3,k−4,6k−5)
⇒(−2+3,2−4,6(2)−5)
⇒(1,−2,7)
⇒‌ Required distance ‌=‌ Distance between points ‌(3,4,4)‌ and ‌(1−2,7)
=√(3−1)2+(4+2)2+(4−7)2
=√4+36+9
=√49
=7‌ units ‌