Find an equation of the line passing through P(0,1,1) and perpendicular to the two lines(d1) [x, y, z] = [1,1,1] + t [0, -1, 2](d2) [x, y, z] = [1,0, -1] + t [2, 1, 1]Select one:a. 0(x+3) + (y-4)+ (z-2) = 0b. [x, y, z] = [-3, 4, 2] + t [0, 1, 1]c. -3x + 4(y-1)+ 2(z-1) = 0d. [x, y, z] = [0, 1, 1] +t [-3, 4, 2]e. None of the other choices is correct

To find the equation of the line passing through P(0,1,1) and perpendicular to the two given lines, we first need to find the direction vector of the new line. This can be done by taking the cross product of the direction vectors of the given lines.

The direction vectors of the given lines are [0, -1, 2] and [2, 1, 1].

The cross product of these vectors is given by:

[-11 - 21, 22 - 01, 0*1 - (-1)*2] = [-2, 4, 2]

So, the direction vector of the new line is [-2, 4, 2].

The equation of a line in 3D space is given by [x, y, z] = [x0, y0, z0] + t [a, b, c], where [x0, y0, z0] is a point on the line and [a, b, c] is the direction vector of the line.

Substituting the given point P(0,1,1) and the direction vector [-2, 4, 2] into this equation, we get:

[x, y, z] = [0, 1, 1] + t [-2, 4, 2]

So, the correct answer is d. [x, y, z] = [0, 1, 1] +t [-2, 4, 2].