If the roots of the quadratic equation $3x^2 - 2x - 1= 0$ are the intercepts of a line, then the line can be

$3x^2 - 2x - 1 = 0$

$3x^2 - 3x + x - 1 = 0$

$3x(x-1) + 1(x-1) = 0$

$(3x+1)(x-1) = 0$

$x = 1$ or $x = -\dfrac{1}{3}$

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The roots are the intercepts of a line, but it is not specified which root is the x-intercept and which is the y-intercept. So both assignments need to be checked using the intercept form of a line:

$\dfrac{x}{a} + \dfrac{y}{b} = 1$

where $a$ = x-intercept and $b$ = y-intercept.

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Taking x-intercept $= 1$ and y-intercept $= -\dfrac{1}{3}$:

$\dfrac{x}{1} + \dfrac{y}{-\frac{1}{3}} = 1$

$x - 3y = 1$

$x - 3y - 1 = 0$ (matches Option A)

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Taking x-intercept $= -\dfrac{1}{3}$ and y-intercept $= 1$:

$\dfrac{x}{-\frac{1}{3}} + \dfrac{y}{1} = 1$

$-3x + y = 1$

$3x - y + 1 = 0$ (matches Option B)

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Since both assignments are valid, the answer is Either (A) or (B).