Let the straight line is y = mx + c

this line passes through the point (-1,2)

So 2 = -m + c

=> c = m+2

Now y = mx+ m+2 .........1

Given equation of line is : x+y = 4 ..........2

Solving equation 1 and 2, we get

x = (2-m)/(m+1) and y = (5m+1)/(m+1)

Now this point is at a distance of 3 unit from the point (-1,2)

From distance formula

√[{(2-m)/(m+1) + 2}² + {(5m+2)/(m+1) - 2}² ]  = 3

Squaring bothe side, we get

      {(2-m)/(m+1) + 1}² + {(5m+2)/(m+1) - 2}² = 9

=> {(2-m+m+1)/(m+1)}² + {(5m+2-2m-2)/(m+1)}² = 9

=> 9/(m+1)² + 9m² /(m+1)² = 0

=> (1+m² )/(m+1)² = 1   (divide by 9 on both side)

=> 1+m² = (m+1)²

=> 1+m² = (m² + 2m +1)

=> 1+m² = m² + 2m + 1

=> 2m =0

=> m =0

Slope of the line is zero.

So the line is parallel to x-axis