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(D) Step $1$: Find the point of intersection of the lines $x + 3y - 1 = 0$ and $x - 2y + 4 = 0$.Subtracting the second equation from the first: $(x +

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$2x - 3y + 7 = 0$

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(D) Step $1$: Find the point of intersection of the lines $x + 3y - 1 = 0$ and $x - 2y + 4 = 0$.Subtracting the second equation from the first: $(x + 3y - 1) - (x - 2y + 4) = 0$ $\Rightarrow 5y - 5 = 0$ $\Rightarrow y = 1$.Substituting $y = 1$ into $x + 3y - 1 = 0$,we get $x + 3(1) - 1 = 0 \Rightarrow x = -2$.So,the point of intersection is $(-2, 1)$.Step $2$: Find the equation of the line perpendicular to $3x + 2y = 0$.The slope of $3x + 2y = 0$ is $m_1 = -3/2$.The slope of the perpendicular line $m_2$ satisfies $m_1 	imes m_2 = -1$,so $m_2 = 2/3$.The equation of the line with slope $2/3$ passing through $(-2, 1)$ is $y - 1 = (2/3)(x + 2)$.Multiplying by $3$: $3y - 3 = 2x + 4 \Rightarrow 2x - 3y + 7 = 0$.

The equation of the straight line perpendicular to the straight line $3x + 2y = 0$ and passing through the point of intersection of the lines $x + 3y - 1 = 0$ and $x - 2y + 4 = 0$ is

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