Find the vector and Cartesian equations of the plane through the points $(1, 2, 3)$ and $(1, 2, 3)$ and perpendicular to the plane $3 x - 2 y + 4 z - 5 = 0$ .

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Solution

The normal vector to the plane $3 x - 2 y + 4 z - 5 = 0$ is $3 \to i - 2 \to j + 4 \to k$
This is parallel to the required plane.
The required plane passes through the points $(1, 2, 3)$ and $(2, 3, 1)$ and parallel to the vector $\to v = 3 \to i - 2 \to j + 4 \to k$ .
The vector equation of the plane is $r = (1 - s) \to a + s \to b + t \to v$
$= (1 - s) (\to i + 2 \to j + 3 \to k) + s (2 \to i + 3 \to j + \to k) + t (3 \to i - 2 \to j + 4 \to k)$
$\Rightarrow (x_{1}, y_{1}, z_{1}) = (1, 2, 3)$
$\Rightarrow (x_{2}, y_{2}, z_{2}) = (2, 3, 1)$
$\Rightarrow 1, m, n = 3, - 2, 4$
Cartesian equation is
$∣ ∣ ∣ ∣ ∣ \begin{matrix} x - x_{1} & y - y_{1} & z - z_{1} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} l & m & n \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$
$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} x - 1 & y - 2 & z - 3 1 & 1 & - 2 3 & - 2 & 4 \end{matrix} ∣ ∣ ∣ ∣ = 0$
$\Rightarrow (x - 1) (4 - 4) - (y - 2) (4 + 6) + (z - 3) (- 2 - 3) = 0$
$\Rightarrow (x - 1) (0) - (y - 2) (10) + (z - 3) (- 5) = 0$
$\Rightarrow - 10 y + 20 - 5 z + 15 = 0$
$\Rightarrow (- 10 y - 5 z + 35 = 0)$ $\div (- 5)$
$\Rightarrow 2 y + z - 7 = 0$

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