Let the plane P : 8x + alpha_1 y + alpha_2 z | vedclass

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(A) Given plane $P : 8x + \alpha_1 y + \alpha_2 z + 12 = 0$ and line $L : \frac{x + 2}{2} = \frac{y - 3}{3} = \frac{z + 4}{5}$.Since the plane $P$ is

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$\sqrt{14}$

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(A) Given plane $P : 8x + \alpha_1 y + \alpha_2 z + 12 = 0$ and line $L : \frac{x + 2}{2} = \frac{y - 3}{3} = \frac{z + 4}{5}$.Since the plane $P$ is parallel to the line $L$,the normal vector of the plane is perpendicular to the direction vector of the line.Thus,$8(2) + \alpha_1(3) + \alpha_2(5) = 0 \Rightarrow 3\alpha_1 + 5\alpha_2 = -16$.Given the $y$-intercept of plane $P$ is $1$,we set $x=0$ and $z=0$ in the plane equation: $\alpha_1(1) + 12 = 0 \Rightarrow \alpha_1 = -12$.Substituting $\alpha_1 = -12$ into $3\alpha_1 + 5\alpha_2 = -16$,we get $3(-12) + 5\alpha_2 = -16 \Rightarrow -36 + 5\alpha_2 = -16 \Rightarrow 5\alpha_2 = 20 \Rightarrow \alpha_2 = 4$.The equation of the plane $P$ is $8x - 12y + 4z + 12 = 0$,which simplifies to $2x - 3y + z + 3 = 0$.The distance from a point on the line $L$ (e.g.,$(-2, 3, -4)$) to the plane $P$ is given by $d = \frac{|2(-2) - 3(3) + 1(-4) + 3|}{\sqrt{2^2 + (-3)^2 + 1^2}}$.$d = \frac{|-4 - 9 - 4 + 3|}{\sqrt{4 + 9 + 1}} = \frac{|-14|}{\sqrt{14}} = \sqrt{14}$.

Let the plane $P : 8x + \alpha_1 y + \alpha_2 z + 12 = 0$ be parallel to the line $L : \frac{x + 2}{2} = \frac{y - 3}{3} = \frac{z + 4}{5}$. If the intercept of $P$ on the $y$-axis is $1$,then the distance between $P$ and $L$ is:

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