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(D) The equation of the line is $4x - 3y + 8 = 0$. The slope of this line is $m_1 = \frac{4}{3}$.Since the perpendicular line passes through $(2, -3)$

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-$2$

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(D) The equation of the line is $4x - 3y + 8 = 0$. The slope of this line is $m_1 = \frac{4}{3}$.Since the perpendicular line passes through $(2, -3)$,its slope is $m_2 = -\frac{1}{m_1} = -\frac{3}{4}$.The equation of the perpendicular line is $y - (-3) = -\frac{3}{4}(x - 2)$,which simplifies to $4y + 12 = -3x + 6$,or $3x + 4y + 6 = 0$.To find the intersection point $M(a, b)$,we solve the system:$4a - 3b = -8$ $(1)$$3a + 4b = -6$ $(2)$Multiplying $(1)$ by $4$ and $(2)$ by $3$: $16a - 12b = -32$ and $9a + 12b = -18$.Adding these gives $25a = -50$,so $a = -2$.Substituting $a = -2$ into $(2)$: $3(-2) + 4b = -6$,so $-6 + 4b = -6$,which gives $b = 0$.Thus,$M(a, b) = (-2, 0)$.We are given $a^3 - b^3 = k^3$,so $(-2)^3 - (0)^3 = k^3$,which means $-8 = k^3$.Therefore,$k = \sqrt[3]{-8} = -2$.

If the perpendicular drawn from the point $(2, -3)$ to the straight line $4x - 3y + 8 = 0$ meets it at $M(a, b)$ and $a^3 - b^3 = k^3$,then $k=$

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