Find the vector equation of the plane passing through points $4 i - 3 j - k, 3 i + 7 j - 10 k$ and $2 i + 5 j - 7 k$ and show that the point $i + 2 j - 3 k$ lies in the plane.

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Solution

Points are (4,-3,-1);(3,7,-10);(2,5,-7)
Equation of plane will be
$∣ ∣ ∣ ∣ \begin{matrix} x - 4 & y + 3 & z + 1 x - 3 & y - 7 & z + 10 x - 2 & y - 5 & z + 7 \end{matrix} ∣ ∣ ∣ ∣ = 0$
$R_{2} \to R_{2} - R_{1} R_{3} \to R_{3} - R_{1}$
$∣ ∣ ∣ ∣ \begin{matrix} x - 4 & y + 3 & z + 1 1 & - 10 & 9 2 & - 8 & 6 \end{matrix} ∣ ∣ ∣ ∣ = 0$
$(x - 4) ∣ ∣ ∣ \begin{matrix} - 10 & 9 - 8 & 6 \end{matrix} ∣ ∣ ∣ - (y + 3) ∣ ∣ ∣ \begin{matrix} 1 & 9 2 & 6 \end{matrix} ∣ ∣ ∣ + (z + 1) ∣ ∣ ∣ \begin{matrix} 1 & - 10 2 & - 8 \end{matrix} ∣ ∣ ∣ = 0$
$(x - 4) 12 - (y + 3) (- 12) + (z + 1) (12) - 0 x - 4 + y + 3 + z + 1 = 0 x + y + z = 0$
For points $(1,, 2, - 3)$ that is $i + 2 j - 3 k$
$1 + 2 - 3 - 0$
$0 = 0$
LHS=RHS
Therefore, $i + 2 j - 3 k$ lies on plane

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