Final answer:
The equation of the straight line passing through the point (2, -1, -1), parallel to the plane 4x+y+z+2=0, and perpendicular to the line x/1=y/-2=z-5/1 is given by the set of parametric equations x = 2 + 3t, y = -1 + 3t, and z = -1 - 9t.
Step-by-step explanation:
To find the equation of a straight line that passes through the point (2, −1, −1) and is parallel to the plane 4x+y+z+2=0 and perpendicular to the line x/1 = y/−2 = (z−5)/1, we need to determine the directional vector of this line. A line that is parallel to a plane is perpendicular to the plane's normal vector, which in this case is (4, 1, 1). Also, because the line is perpendicular to another line with a direction vector given by (1, −2, 1), the two vectors should be orthogonal. Thus the direction vector of the required line is the cross product of the plane's normal vector and the given line's direction vector.
Let's compute the cross product of (4, 1, 1) and (1, −2, 1):
λ = 4i + j + k
μ = i − 2j + k
The cross product λ x μ = (3, 3, −9).
This gives us the direction vector for the line we seek. Now we can form the equation of the line using the point-slope form:
−(x − 2)/3 = (y + 1)/3 = −(z + 1)/9
Therefore, the equation of the straight line is:
x = 2 + 3t
y = −1 + 3t
z = −1 − 9t