Determine if the point (4,-2,3) lies in the plane with vector equation (x, y, z) = (2, 0, -1) + s(4, -2, 1) + t(-3, -1, 2).

#1: 4s - 3t +2 = 4 ----> 4s - 3t = 2

#2: -2s - t + 0 = -2 ----> t = 2 - 2s

using substitution:
4s - 3(2-2s) = 2
4s - 6 + 6s = 2
10s = 8
s = 4/5
then t = 2 - 8/5 = 2/5

#3: is s + 2t - 1 = 3 ???
LS=4/5 + 4/5 - 1
= 3/5
RS = 3
NOPE, given point is not on the plane

To determine if the point (4, -2, 3) lies in the plane with vector equation (x, y, z) = (2, 0, -1) + s(4, -2, 1) + t(-3, -1, 2), we can substitute the coordinates of the point into the equation and solve for the values of s and t.

Substituting the given coordinates (4, -2, 3) into the equation, we have:

x = 2 + 4s - 3t
y = 0 - 2s - t
z = -1 + s + 2t

Now let's substitute the values of x, y, and z with their corresponding values from the point:

4 = 2 + 4s - 3t (equation 1)
-2 = -2s - t (equation 2)
3 = -1 + s + 2t (equation 3)

We have a system of three equations with two unknowns, s and t. To solve for s and t, we need to eliminate one variable. We can start by isolating s in equation 2:

-2 = -2s - t
2s = -t - 2
s = (-t - 2)/2
s = -t/2 - 1 (equation 4)

Now substitute the value of s in equations 1 and 3:

4 = 2 + 4s - 3t
4 = 2 + 4(-t/2 - 1) - 3t
4 = 2 - 2t - 4 - 3t
4 = -5t - 2
5t = -6
t = -6/5 (equation 5)

Substitute the value of t in equation 4:

s = -t/2 - 1
s = -(-6/5)/2 - 1
s = 3/5 - 1
s = -2/5 (equation 6)

Now we have the values of s and t. To check if the point (4, -2, 3) lies in the plane, we need to substitute the calculated values of s and t into the original equation and see if it is satisfied.

(x, y, z) = (2, 0, -1) + s(4, -2, 1) + t(-3, -1, 2)

Substituting s = -2/5 and t = -6/5:

(x, y, z) = (2, 0, -1) + (-2/5)(4, -2, 1) + (-6/5)(-3, -1, 2)
(x, y, z) = (2, 0, -1) + (-8/5, 4/5, -2/5) + (18/5, 6/5, -12/5)
(x, y, z) = (2 - 8/5 + 18/5, 0 + 4/5 + 6/5, -1 - 2/5 - 12/5)
(x, y, z) = (10/5, 10/5, -20/5)
(x, y, z) = (2, 2, -4)

The coordinates (4, -2, 3) do not satisfy the equation when we substitute the values of s and t. Therefore, the point (4, -2, 3) does not lie in the plane with the given vector equation.

To determine if the point (4, -2, 3) lies in the plane with vector equation (x, y, z) = (2, 0, -1) + s(4, -2, 1) + t(-3, -1, 2), we need to check if there exist values of s and t that satisfy the equation.

First, we need to find the coordinates of a point in the plane. From the vector equation, we can see that the point (2, 0, -1) lies in the plane.

Next, we can express the equation of the plane in Cartesian form. Let's call the normal vector of the plane N = (A, B, C), and let P = (x0, y0, z0) be a point in the plane. The equation of the plane can be written as:

A(x - x0) + B(y - y0) + C(z - z0) = 0.

In our case, the normal vector is (4, -2, 1) cross (-3, -1, 2). To find the normal vector, we take the cross product of the two direction vectors in the equation. Thus, the normal vector is:

N = (4, -2, 1) cross (-3, -1, 2) = (3, -5, -2).

Now, we have a point P = (2, 0, -1) in the plane, and the normal vector N = (3, -5, -2).

Using the Cartesian form of the equation of the plane, we can substitute the coordinates of the given point (4, -2, 3) into the equation:

3(x - 2) - 5(y - 0) - 2(z + 1) = 0.

Simplifying this equation, we get:

3x - 6 - 5y - 2z - 2 = 0.

3x - 5y - 2z = 8.

Now, we can substitute the coordinates of the given point (4, -2, 3) into the equation:

3(4) - 5(-2) - 2(3) = 12 + 10 - 6 = 16.

Since 16 is not equal to 8, the point (4, -2, 3) does not lie in the plane.

Therefore, the answer is no, the point (4, -2, 3) does not lie in the plane with the given vector equation.