If the angle between the planes ax - y + 3z = | vedclass

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(A) The normal vectors to the planes $ax - y + 3z = 2a$ and $3x + ay + z = 3a$ are $\vec{n_1} = (a, -1, 3)$ and $\vec{n_2} = (3, a, 1)$ respectively.G

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$(2, -1, 2)$

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(A) The normal vectors to the planes $ax - y + 3z = 2a$ and $3x + ay + z = 3a$ are $\vec{n_1} = (a, -1, 3)$ and $\vec{n_2} = (3, a, 1)$ respectively.Given the angle $	heta = \frac{\pi}{3}$ between the planes, we have $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.$\cos(\frac{\pi}{3}) = \frac{|3a - a + 3|}{\sqrt{a^2 + (-1)^2 + 3^2} \sqrt{3^2 + a^2 + 1^2}} = \frac{1}{2}$.$\frac{|2a + 3|}{\sqrt{a^2 + 10} \sqrt{a^2 + 10}} = \frac{1}{2} \implies \frac{|2a + 3|}{a^2 + 10} = \frac{1}{2}$.$2|2a + 3| = a^2 + 10$.Case $1$: $4a + 6 = a^2 + 10 \implies a^2 - 4a + 4 = 0 \implies (a - 2)^2 = 0 \implies a = 2$.Case $2$: $-4a - 6 = a^2 + 10 \implies a^2 + 4a + 16 = 0$, which has no real roots.Thus, $a = 2$.The plane equation becomes $(2+2)x + (2-4)y + 2(2)z = 2$, which is $4x - 2y + 4z = 2$, or $2x - y + 2z = 1$.The direction ratios of the normal to this plane are $(2, -1, 2)$.

If the angle between the planes $ax - y + 3z = 2a$ and $3x + ay + z = 3a$ is $\frac{\pi}{3}$, then the direction ratios of the line perpendicular to the plane $(a+2)x + (a-4)y + 2az = a$ are

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