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The equations
$(λ - 1) x + (3 λ + 1) y + 2 λ z = 0,$
$(λ - 1) x + (4 λ - 2) y + (λ + 3) z = 0,$
and $2 x + (3 λ + 1) y + 3 (λ - 1) z = 0$
give non – trivial solution for some values of $λ$ then the ratio $x : y : z$ is -----
( when $λ$ has smallest of these values)

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3 : 2 : 1

1 : 3 : 2

1 : 3 : 1

1 : 1 : 1

answer is D.

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Detailed Solution

$| \begin{matrix} λ - 1 & 3 λ + 1 & 2 λ \\ λ - 1 & 4 λ - 2 & λ + 3 \\ 2 & 3 λ + 1 & 3 (λ - 1) \end{matrix} | = 0$

$\Rightarrow λ = 0 o r 3$

$- x - 2 y + 3 z = 0 a n d 2 x + y - 3 z = 0,$

$∴ \frac{x}{6 - 3} = \frac{y}{6 - 3} = \frac{z}{- 1 + 4} \Rightarrow \frac{x}{1} = \frac{y}{1} = \frac{z}{1}$

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