Let pi_1 be the plane passing through the poi | vedclass

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(D) The normal vectors to the planes $\pi_1$ and $\pi_2$ are $\vec{n}_1 = a\hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{n}_2 = \hat{i} - 2\hat{j} + \hat{k

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$1$

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(D) The normal vectors to the planes $\pi_1$ and $\pi_2$ are $\vec{n}_1 = a\hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{n}_2 = \hat{i} - 2\hat{j} + \hat{k}$ respectively.The angle $	heta$ between two planes is given by $\cos 	heta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{||\vec{n}_1|| ||\vec{n}_2||}$.Given $\cos 	heta = -\sqrt{\frac{3}{7}}$, we consider the magnitude of the cosine of the angle between the normals: $\cos^2 	heta = \frac{3}{7}$.$\vec{n}_1 \cdot \vec{n}_2 = (a)(1) + (2)(-2) + (-3)(1) = a - 4 - 3 = a - 7$.$||\vec{n}_1|| = \sqrt{a^2 + 2^2 + (-3)^2} = \sqrt{a^2 + 13}$.$||\vec{n}_2|| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$.Thus, $\frac{(a-7)^2}{(a^2+13)(6)} = \frac{3}{7}$.$7(a^2 - 14a + 49) = 18(a^2 + 13)$.$7a^2 - 98a + 343 = 18a^2 + 234$.$11a^2 + 98a - 109 = 0$.$(a-1)(11a+109) = 0$.Since $a$ is an integer, $a = 1$.

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