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

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Solution

$\begin{matrix} D = (6, 5, - 1) L e t - - \to A B = [4 - 3, x - 2, 5 - 2] = [1, x - 2, 4] - - \to A C = [1, 0, - 3] - - \to A D = [3, 3, - 2] G i v e n p o i n t s a r e c o - p l a n e r - - \to A B . (- - \to A C \times - - \to A D) = 0 ∣ ∣ ∣ ∣ \begin{matrix} 1 & x - 2 & 4 1 & 0 & - 3 3 & 3 & - 2 \end{matrix} ∣ ∣ ∣ ∣ = 0 E x p a n d i n g b y C_{1} 1 (0 + 9) - 1 (- 2 (x - 2) - 12) + 3 (- 3 (x - 2)) = 0 9 - 1 (- 2 x + 4 - 12) + 3 (- 3 x + 6) = 0 9 + 2 x + 8 - 9 x + 18 = 0 - 7 x + 35 = 0 - 7 x = - 35 x = \frac{35}{7} ∴ x = 5 H e n c e, t h e v a l u e o f x i s 5. \end{matrix}$

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