Determine the value of k such that the points (4,-2,6), B(0,1,0) and C(1,0,-5) and D (1,k,-2) lie on the same plane.

(B-A)x(C-A) = {-4,3,-6}x{-3,2,-11} = {-21,-26,1}

If D is in the plane of ABC, then (D-A)x(C-A) will also be normal to the plane of ABC. That is,

{-3,k+2,-8}x{-3,2,-11} = {-21,-26,1}
The plane is
21x+26y-z = 26

So,

21+26k+2 = 26
26k = 3
k = 3/26

To determine the value of k such that the points (4, -2, 6), B(0, 1, 0), C(1, 0, -5), and D(1, k, -2) lie on the same plane, we can use the concept of vectors.

Step 1: Find two vectors that lie on the plane formed by the given points. We can select vectors AB and AC.

Vector AB = B - A = (0, 1, 0) - (4, -2, 6) = (-4, 3, -6)
Vector AC = C - A = (1, 0, -5) - (4, -2, 6) = (-3, 2, -11)

Step 2: Calculate the cross product of vectors AB and AC. The cross product of two vectors gives us a normal vector to the plane.

Normal vector N = AB x AC
= (-4, 3, -6) x (-3, 2, -11)
= (-33, -2, -14)

Step 3: Now we can use the equation of the plane, which is given by the dot product of the normal vector with any point on the plane, to find k.

Let's consider the point D(1, k, -2).

Dot product of N and D = (-33, -2, -14) . (1, k, -2)
= -33(1) + (-2)(k) + (-14)(-2)
= -33 - 2k + 28
= -5 - 2k

Since the points lie on the same plane, the dot product of N and D should be equal to 0.

-5 - 2k = 0
-2k = 5
k = -5/2
k = -2.5

Therefore, the value of k that makes the points (4, -2, 6), B(0, 1, 0), C(1, 0, -5), and D(1, k, -2) lie on the same plane is k = -2.5.

To determine the value of k such that the points (4,-2,6), B(0,1,0), C(1,0,-5), and D(1,k,-2) lie on the same plane, we need to check if the vectors formed by these points lie on the same plane.

To do this, we can calculate the normal vector of the plane formed by the points B, C, and D. If the normal vector is perpendicular to a vector drawn from any point on the plane to the fourth point (4,-2,6), then all four points lie on the same plane.

Step 1: Find the vectors formed by the points B, C, and D.
The vector formed by points B and C is given by:
BC = C - B = (1, 0, -5) - (0, 1, 0) = (1, -1, -5).

The vector formed by points B and D is given by:
BD = D - B = (1, k, -2) - (0, 1, 0) = (1, k - 1, -2).

Step 2: Calculate the normal vector to the plane formed by B, C, and D.
The normal vector of a plane formed by three non-collinear points can be found using their corresponding vectors:

N = BC x BD

where "x" represents the cross product.

Calculating the cross product:
N = (1, -1, -5) x (1, k - 1, -2)

N = [(-1)(-2) - (-1)(k - 1), (-1)(1) - (-5)(1), (1)(k - 1) - (-2)(-1)]

N = [2 + k - 1, -1 + 5, k - 1 + 2]

N = [k + 1, 4, k + 1].

Step 3: Check if the normal vector is perpendicular to a vector from any point on the plane to the fourth point (4,-2,6).
To determine if the vectors are perpendicular, we need to find their dot product.

The dot product of the normal vector and the vector (4,-2,6) - B, where B is any point on the plane, should be zero.

Let's take B as the point (0,1,0):
(4, -2, 6) - (0, 1, 0) = (4, -3, 6)

Now, calculate the dot product:
N · (4, -3, 6) = (k + 1)(4) + 4(6) + (k + 1)(-3) = 4k + 4 + 24 - 3k - 3 = k + 25.

For the points (4,-2,6), B(0,1,0), C(1,0,-5), and D(1,k,-2) to lie on the same plane, the dot product should be zero:

k + 25 = 0.

Solving for k:
k = -25.

Therefore, the value of k that makes the four points lie on the same plane is k = -25.