If three vectors $x \vec{a} - 2 \vec{b} + 3 \vec{c}$ , $- 2 \vec{a} + y \vec{b} - 4 \vec{c}$ and $- z \vec{b} + 2 \vec{c}$ are coplanar, where $\vec{a}$ , $\vec{b}$ and $\vec{c}$ are unit (or any) vectors, then:

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$x y - 2 z x + 3 z - 4 = 0$

$4 x y - 3 z x + 3 z = 4$

$2 x y - 2 z x - 3 z - 4 = 0$

$x y + 3 z x - 3 x = 4$

answer is D.

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Detailed Solution

Conditions of coplanarity, coplanar vectors have scalar triple product as zero

$\vec{A} . (\vec{B} \times \vec{C}) = 0$

or $| \begin{matrix} x & - 2 & 3 \\ - 2 & y & - 4 \\ 0 & - z & 2 \end{matrix} | = 0$

or $x (2 y - 4 z) + 2 (- 4 - 0) + 3 (2 z - 0) = 0$

or $2 x y - 4 z x - 8 + 6 z = 0$

or $x y - 2 z x + 3 z - 4 = 0$

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