Find the equation of the plane passing through the points $(1, 2, 3), (3, 5, 7)$ and $(4, 3, 1)$

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Solution

Let $A (1, 2, 3); B (3, 5, 7); C (4, 3, 1)$
Three points (A,B,C) can define two distinct vectors AB and AC. Since the two vectors lie on the plane, their cross product can be used as a normal to the plane.
Determine the vectors
Find the cross product of the two vectors
Substitute one point into the Cartesian equation to solve for d.
$- - \to A B = (x_{B} - x_{A})^i + (y_{B} - y_{A})^j + (z_{B} - z_{A})^k$ $⟹ - - \to A B = 2^i + 3^j + 4^k - - \to A C = (x_{c} - x_{A})^i + (y_{c} - y_{A})^j + (z_{c} - z_{A})^k ⟹ - - \to A C = 3^i + 1^j - 2^k - - \to A B \times - - \to A C = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 2 & 3 & 4 3 & 1 & - 2 \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 10^i + 14^j - 7^k$
The equation of the plane is $- 10 x + 14 y - 7 z = d$
Plug any point in the equation to fing $d$
$⟹ - 10 (1) + 14 (2) - 7 (3) = d ⟹ - 3 = d$
The equation of the plane is $- 10 x + 14 y - 7 z = - 3$

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