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

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Solution

Four points $A, B, C, D$ are coplanar if the three vectors $\to A B, \to A C,$ and $\to A D$ are coplanar.
$[\to A B, \to A C, \to A D] = 0$
$\to A B = \to O B - \to O A$
$= (4 - 3)^i + (x - 2)^j + (5 - 1)^k$
$=^i + (x - 2)^j + 4^k$
$\to A C = \to O C - \to O A$
$= (4 - 3)^i + (2 - 2)^j + (- 2 - 1)^k$
$=^i - 3^k$
$\to A D = \to O D - \to O A$
$= (6 - 3)^i + (5 - 2)^j + (- 1 - 1)^k$
$= 3^i + 3^j - 2^k$
$[\to A B, \to A C, \to A D] = ∣ ∣ ∣ ∣ \begin{matrix} 1 & x - 2 & 4 1 & 0 & - 3 3 & 3 & - 2 \end{matrix} ∣ ∣ ∣ ∣$
$\Rightarrow 0 = 1 [0 - (- 9)] - (x - 2) [- 2 - (- 9)] + 4 [3 - 0]$
$\Rightarrow 0 = 9 - (x - 2) 7 + 12$
$\Rightarrow 35 - 7 x = 0$
$\Rightarrow x = \frac{35}{7} = 5$

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