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(B) Let the required line pass through the point $P(0,1,2)$ with direction ratios $(a, b, c)$.The equation of the line is $\frac{x-0}{a} = \frac{y-1}{

Vedclass · Accepted Answer

$\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}$

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(B) Let the required line pass through the point $P(0,1,2)$ with direction ratios $(a, b, c)$.The equation of the line is $\frac{x-0}{a} = \frac{y-1}{b} = \frac{z-2}{c}$.The given line is $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{-2}$,which has direction ratios $(2, 3, -2)$.Since the two lines are perpendicular,the dot product of their direction ratios must be zero:$2a + 3b - 2c = 0$.Checking the options:For option $C$: The direction ratios are $(3, 4, 3)$.$2(3) + 3(4) - 2(3) = 6 + 12 - 6 = 12 
eq 0$.For option $B$: The direction ratios are $(-3, 4, 3)$.$2(-3) + 3(4) - 2(3) = -6 + 12 - 6 = 0$.Thus,the line $\frac{x}{-3} = \frac{y-1}{4} = \frac{z-2}{3}$ is perpendicular to the given line.

The equation of the line passing through the point $(0,1,2)$ and perpendicular to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}$ is

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