a, b, c are three non-coplanar vector. Show that the points with $p . v^{'} s$
$6 ¯ ¯ ¯ a - 4 ¯ ¯ b + 10 ¯ ¯ c, - 5 ¯ ¯ ¯ a + 3 ¯ ¯ b + 10 ¯ ¯¯ ¯ c, 4 ¯ ¯ ¯ a - 6 ¯ ¯ b - 10 ¯ ¯ c$ and $2 ¯ ¯ b + 10 ¯ ¯ c$ are coplanar.

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Solution

Let $\to p = 6 \to a - 4 \to b + 10 \to c (P)$
$\to q = - 5 \to a + 3 \to b + 10 \to c (Q)$
$\to r = 4 \to a - 6 \to b - 10 \to c (R)$
$\to s = 2 \to b + 10 \to c$
$P Q = - 11 \to a + r \to b$
$P R = - 2 \to a - 2 \to b - 20 \to c$
$P S = - 6 \to a + 6 \to b$
$[P Q P R P S] = 0$ for coplarality of points
$[P Q P R P S = ∣ ∣ ∣ ∣ \begin{matrix} - 11 & 7 & 0 - 2 & - 2 & - 20 - 6 & 6 & 0 \end{matrix} ∣ ∣ ∣ ∣ [\to a \to b \to c]$
$= 20 (- 66 + 42) [\to a \to b \to c]$
$= - 240 [\to a \to b \to c]$
$[\to a \to b \to c] \neq 0 \to$ not coplanar
So there points are not coplanar

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