Jawaban:

(a) The position vector of point O is given by O = 0i + 0j + 0k. The position vector of point C can be calculated as the sum of vectors AB and BO:

C = A + B = (i + j) + (-2i + k) = -i + j + k

So, the vector equation of line OC is given by

OC = C - O = (-i + j + k) - (0i + 0j + 0k) = -i + j + k.

(b) To find the coordinates of P, we need to find the point that minimizes the distance between PA and PB. The shortest distance between two points in 3-dimensional space is given by the distance formula:

d = √((x2-x1)² + (y2-y1)² + (z2-z1)²)

where (x1, y1, z1) and (x2, y2, z2) are the coordinates of two points.

We can use this formula to calculate the distances PA and PB, then find the point that minimizes the sum of the two distances.

The coordinates of A and B are (1, 1, 0) and (-2, 0, 1), respectively. Let's assume that the coordinates of P are (x, y, z). Then the distances PA and PB are given by

PA = √((x - 1)² + (y - 1)² + (z - 0)²)

PB = √((x + 2)² + (y - 0)² + (z - 1)²)

Since PA PB must have the minimum value, the derivative of the sum of these distances must equal 0.

Taking the partial derivative of PA PB with respect to x, y, and z and setting them equal to 0, we obtain:

∂(PA + PB)/∂x = 0 = 2x + 4

∂(PA + PB)/∂y = 0 = 2y + 2

∂(PA + PB)/∂z = 0 = 2z

Solving these equations simultaneously, we get:

x = -2, y = -1, z = 0

So, the coordinates of point P that minimize the sum of the distances PA and PB are (-2, -1, 0)