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The given equation of the line is $\frac{2x+4}{6}=\frac{y+1}{2}=\frac{-2z+6}{-4}$. We can rewrite this as:\r\n\r\n$\frac{2(x+2)}{6}=\frac{y+1}{2}=\frac{-2(z-3)}{-4}$ which simplifies to $\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$.
Let the common ratio be $\lambda$. Then, any point $Q$ on the line can be expressed as:\r\n\r\n$x = 3\lambda - 2$\r\n$y = 2\lambda - 1$\r\n$z = 2\lambda + 3$\r\n\r\nSo, the coordinates of point $Q$ are $(3\lambda - 2, 2\lambda - 1, 2\lambda + 3)$.
The distance between $P(1,2,3)$ and $Q(3\lambda - 2, 2\lambda - 1, 2\lambda + 3)$ is given as $3\sqrt{2}$. Using the distance formula:\r\n\r\n$\sqrt{((3\lambda - 2) - 1)^2 + ((2\lambda - 1) - 2)^2 + ((2\lambda + 3) - 3)^2} = 3\sqrt{2}$\r\n\r\n$\sqrt{(3\lambda - 3)^2 + (2\lambda - 3)^2 + (2\lambda)^2} = 3\sqrt{2}$
Squaring both sides, we get:\r\n\r\n$(3\lambda - 3)^2 + (2\lambda - 3)^2 + (2\lambda)^2 = (3\sqrt{2})^2$\r\n\r\n$9\lambda^2 - 18\lambda + 9 + 4\lambda^2 - 12\lambda + 9 + 4\lambda^2 = 18$\r\n\r\n$17\lambda^2 - 30\lambda + 18 = 18$\r\n\r\n$17\lambda^2 - 30\lambda = 0$\r\n\r\n$\lambda(17\lambda - 30) = 0$
From the above equation, we have two possible values for $\lambda$:\r\n\r\n$\lambda = 0$ or $17\lambda - 30 = 0 \implies \lambda = \frac{30}{17}$
Case 1: $\lambda = 0$\r\n\r\n$x = 3(0) - 2 = -2$\r\n$y = 2(0) - 1 = -1$\r\n$z = 2(0) + 3 = 3$\r\n\r\nSo, $Q_1 = (-2, -1, 3)$\r\n\r\nCase 2: $\lambda = \frac{30}{17}$\r\n\r\n$x = 3(\frac{30}{17}) - 2 = \frac{90}{17} - \frac{34}{17} = \frac{56}{17}$\r\n$y = 2(\frac{30}{17}) - 1 = \frac{60}{17} - \frac{17}{17} = \frac{43}{17}$\r\n$z = 2(\frac{30}{17}) + 3 = \frac{60}{17} + \frac{51}{17} = \frac{111}{17}$\r\n\r\nSo, $Q_2 = (\frac{56}{17}, \frac{43}{17}, \frac{111}{17})$
The points $Q$ are $(-2, -1, 3)$ and $(\frac{56}{17}, \frac{43}{17}, \frac{111}{17})$.
\r\n Final Answer: (-2, -1, 3) and (56/17, 43/17, 111/17)\r\n <\/p>\r\n <\/div>\r\n <\/div>
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