One way to solve this kind of problem is to use a description of lines drawn from formal linear algebra.
The line (which will be referred to as Line1): y = x +7 can be described by the ordered triplet (x,y,z)
(0, 7, 0) + α (1 , 1 ,0) = ( α , α +7, 0) The parameter α can take on all real values. It is the parameter that moves a point along the line.
Line 1 is along the 45 degree direction in the x,y plane. The vector (1, 1, 0) above generates a 45 degree line in the x,y plane. A similar vector, namely (1, -1, c) will generate a line in the 3-space that will either be skew to Line1 or intersect line Line1 at right angles. This is because the dot product between (1,1,0) and (1, -1, c) is zero.
To get the intersection case for Line2 we can write for Line2
( α , α +7, 0) + β (1, -1, c) The parameter β plays a role for Line2, similar to that of α for Line1
This expression for Line2 is actually a family of lines indexed by α. The desired line is one member of this family.
The next step is to set ( α , α +7, 0) + β (1, -1, c) = (-4, -4, -21) This can be viewed as a set of three equations for three unknowns (α, β, c ). The equations are:
α + β = -4
α + 7 - β = -4
β c = -21
The first two equations can be solved to get α = -15/2 β = 7/2
The third equation then gives c = -6
Substituting for α and c gives for Line2 (-15/2 + β , -1/2 - β , -6 β)
For β = 7/2 this form generates (-4 , -4, -21) as required. Also with β =0 we recover ( α , α +7, 0) when α =-15/2 . These two facts satisfy the requirements for Line2: It passes through (-4,-4,-21) and lies in a plane intersecting the x,y plane along the line y = x +7 .
From the viewpoint of linear algebra, the form (-15/2 + β , -1/2 - β , -6 β) is a complete answer. It can be converted to alternative forms. For example eliminating β in favor of z results in the set of two equations
x = -15/2 - z/6 and
y = -1/2 + z/6