Find the equation of the line passing through | vedclass

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(D) Let the given line be $L: \frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{-1} = k$. Any point $Q$ on $L$ is $(2k - 1, 3k + 3, -k - 2)$.The direct

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$\frac{x - 2}{-24} = \frac{y + 3}{13} = \frac{z - 1}{-9}$

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(D) Let the given line be $L: \frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{-1} = k$. Any point $Q$ on $L$ is $(2k - 1, 3k + 3, -k - 2)$.The direction ratios of $PQ$ are $(2k - 1 - 2, 3k + 3 - (-3), -k - 2 - 1) = (2k - 3, 3k + 6, -k - 3)$.Since $PQ \perp L$, the dot product of the direction ratios of $PQ$ and $L$ is zero:$2(2k - 3) + 3(3k + 6) - 1(-k - 3) = 0$.$4k - 6 + 9k + 18 + k + 3 = 0 \implies 14k + 15 = 0 \implies k = -\frac{15}{14}$.The direction ratios of $PQ$ are $(2(-\frac{15}{14}) - 3, 3(-\frac{15}{14}) + 6, -(-\frac{15}{14}) - 3) = (-\frac{72}{14}, \frac{39}{14}, -\frac{27}{14})$.Multiplying by $-\frac{14}{3}$, we get the direction ratios as $(24, -13, 9)$.Wait, checking the dot product again: $2(24) + 3(-13) - 1(9) = 48 - 39 - 9 = 0$. The vector is $(24, -13, 9)$.The line equation is $\frac{x - 2}{24} = \frac{y + 3}{-13} = \frac{z - 1}{9}$.Comparing with options, option $A$ is $\frac{x - 2}{24} = \frac{y + 3}{13} = \frac{z - 1}{9}$. Re-evaluating the direction ratios: $(2k-3, 3k+6, -k-3)$. For $k=-15/14$, $DRs = (-72/14, 39/14, -27/14) \propto (-24, 13, -9)$.Thus, the equation is $\frac{x - 2}{-24} = \frac{y + 3}{13} = \frac{z - 1}{-9}$, which matches option $D$.

Find the equation of the line passing through the point $P(2, -3, 1)$ and perpendicular to the line $L: \frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{-1}$.

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