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(B) Step $1$: Find the point of intersection of $x + 2y + 6 = 0$ and $2x - y = 2$.From the second equation, $y = 2x - 2$.Substitute into the first: $x

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$39x - 2y + 10 = 0$

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(B) Step $1$: Find the point of intersection of $x + 2y + 6 = 0$ and $2x - y = 2$.From the second equation, $y = 2x - 2$.Substitute into the first: $x + 2(2x - 2) + 6 = 0 \implies x + 4x - 4 + 6 = 0 \implies 5x + 2 = 0 \implies x = -2/5$.Then $y = 2(-2/5) - 2 = -4/5 - 10/5 = -14/5$.The point of intersection is $(-2/5, -14/5)$.Step $2$: Use the family of lines equation: $(x + 2y + 6) + \lambda(2x - y - 2) = 0$.This simplifies to $(1 + 2\lambda)x + (2 - \lambda)y + (6 - 2\lambda) = 0$.Step $3$: The $y$-intercept is $5$, so the line passes through $(0, 5)$.Substitute $(0, 5)$ into the equation: $(1 + 2\lambda)(0) + (2 - \lambda)(5) + (6 - 2\lambda) = 0$.$10 - 5\lambda + 6 - 2\lambda = 0 \implies 16 - 7\lambda = 0 \implies \lambda = 16/7$.Step $4$: Substitute $\lambda = 16/7$ back into the equation:$(1 + 2(16/7))x + (2 - 16/7)y + (6 - 2(16/7)) = 0$.$(7 + 32)/7 x + (14 - 16)/7 y + (42 - 32)/7 = 0$.$39x - 2y + 10 = 0$.

The equation of the line passing through the point of intersection of the lines $x + 2y + 6 = 0$ and $2x - y = 2$ and making an intercept $5$ on the $y$-axis is

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