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(D) Given $|a| = \sqrt{3}$, $|b| = 3$, and $|a - b| = 4$. Squaring $|a - b| = 4$ gives $|a|^2 + |b|^2 - 2(a \cdot b) = 16$, so $3 + 9 - 2(a \cdot b) =

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$128$

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(D) Given $|a| = \sqrt{3}$, $|b| = 3$, and $|a - b| = 4$. Squaring $|a - b| = 4$ gives $|a|^2 + |b|^2 - 2(a \cdot b) = 16$, so $3 + 9 - 2(a \cdot b) = 16$, which implies $a \cdot b = -2$.Let $v = 2i + 2j - k$. The equation $a 	imes v = v 	imes b$ can be written as $a 	imes v + b 	imes v = 0$, so $(a + b) 	imes v = 0$.This implies $(a + b)$ is parallel to $v = 2i + 2j - k$. Thus, $a + b = k(2i + 2j - k)$ for some scalar $k$.Calculate $|a + b|^2 = |a|^2 + |b|^2 + 2(a \cdot b) = 3 + 9 + 2(-2) = 8$.Since $|a + b|^2 = k^2(2^2 + 2^2 + (-1)^2) = 9k^2$, we have $9k^2 = 8$, so $k^2 = 8/9$ and $|k| = \frac{2\sqrt{2}}{3}$.We need $|(a + b) \cdot (3i + 4j + 2k)| = |k(2i + 2j - k) \cdot (3i + 4j + 2k)| = |k(6 + 8 - 2)| = |12k| = 12 \cdot \frac{2\sqrt{2}}{3} = 8\sqrt{2}$.Given this equals $\sqrt{\lambda}$, so $\sqrt{\lambda} = 8\sqrt{2} = \sqrt{64 \cdot 2} = \sqrt{128}$. Thus $\lambda = 128$.

Let $a$ and $b$ be linearly independent vectors such that $|a| = \sqrt{3}$, $|b| = 3$, and $|a - b| = 4$. If $a \times (2i + 2j - k) = (2i + 2j - k) \times b$ and $|(a + b) \cdot (3i + 4j + 2k)| = \sqrt{\lambda}$, then $\lambda = ...$

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