Representing the following line using the symmetric form and parametric equations {█(x+y+z+1=0@2x-y+4z+4=0)┤ .

To find the symmetric form of the line, we can write each equation in terms of a normal vector and a distance from the origin.

For the first equation, the normal vector is <1, 1, 1> and the distance is -1 (since the equation passes through the point (-1, 0, 0) which is one unit away from the origin in the direction of the normal vector).

Similarly, for the second equation, the normal vector is <2, -1, 4> and the distance is -4 (since the equation passes through the point (-2, 0, 1) which is four units away from the origin in the direction of the normal vector).

Using this information, we can write the symmetric form of the line as:

(x-1)/1 = (y-0)/1 = (z-0)/1 = -(d/|n|)

where d is the distance from the origin and n is the normal vector.

Simplifying, we get:

(x-1)/1 = (y-0)/1 = (z-0)/1 = -(1/√3)

So the symmetric form of the line is:

x-1 = y = z

and

-(x-1)/√3 = -(y-0)/√3 = -(z-0)/√3

To find the parametric equations of the line, we can choose any two points on the line and write them in vector form.

Let's choose the points (-1, 0, 0) and (-2, 1, -1). Then the direction vector of the line is:

< -1 - (-2), 0 - 1, 0 - (-1) > = <1, -1, 1>

So the parametric equations of the line are:

x = -1 + t
y = t
z = -1 + t

where t is any real number.