Find the value of x such that the points A(3, 2, 1), B(4, x, 5), C(4, 2, - 2) and D(6, 5, - 1) are coplanar.

We will solve this question by using planes method, first we will find the general equation of the plane containing all these points starting with the first point A,

a(x - 3) + b(y - 2) + c(z - 1) = 0

Now putting point C,

a(4 - 3) + b(2 - 2) + c( - 2 - 1) = 0

a + 0b – 3c = 0

Now putting point D,

a(6 - 3) + b(5 - 2) + c( - 1 - 1) = 0

3a + 3b - 2c = 0

Solving these equations by cross - multiplication method, we get,

a = 9λ, b = - 7λ, c = 3λ

The equation of the plane is,

9(x - 3) - 7(y - 2) + 3(z - 1) = 0

Putting point B in the equation,

9(4 - 3) - 7(x - 2) + 3(5 - 1) = 0

9 – 7x + 14 + 12 = 0

x = 5