The position vector of A B and C are given as vector a=xi-4j-2k vector b =-5i+yj+k and vector c=10i-8j+2k respectively where xyz are constant given that A B C are collinear and that 3AC=-2AB find xyz?

To determine the values of x, y, and z, we need to use the given information that A, B, and C are collinear and that 3AC = -2AB.

Let's start by finding the position vectors of A, B, and C using the given information:

Vector a = xi - 4j - 2k
Vector b = -5i + yj + k
Vector c = 10i - 8j + 2k

Since A, B, and C are collinear, we can write the following relationship:

AB = k * AC

where k is a scalar.

Now, let's find the vectors AB and AC:

AB = Vector B - Vector A
= (-5i + yj + k) - (xi - 4j - 2k)
= -5i + yj + k - xi + 4j + 2k
= (-5 - x)i + (y + 4)j + (1 + k)k

AC = Vector C - Vector A
= (10i - 8j + 2k) - (xi - 4j - 2k)
= 10i - 8j + 2k - xi + 4j + 2k
= (10 - x)i + (-8 + 4)j + (2 + 2)k
= (10 - x)i - 4j + 4k

Now, we are given that 3AC = -2AB. Let's set up this equation:

3(10 - x)i - 12j + 12k = -2(-5 - x)i + 2(y + 4)j + 2(1 + k)k

Simplifying this equation, we get:

30 - 3x = 10 + 2x
-3x - 2x = 10 - 30
-5x = -20
x = 4

Now that we have the value of x, we can substitute it back into the equation to find y and z:

3(10 - 4) - 12 = -2(-5 - 4) + 2(y + 4) + 2(1 + z)

18 - 12 = 2(9) + 2(y + 4) + 2(1 + z)
6 = 18 + 2(y + 4) + 2(1 + z)
6 = 18 + 2y + 8 + 2z + 2
6 = 28 + 2y + 2z
-22 = 2y + 2z
-11 = y + z

Therefore, the values of x, y, and z are x = 4, y = -11 - z, and z can be any constant.