If $A = (1, 2, 1), B = (1, 4, 6), C = (5, 4, 4)$ are three points then the foot of the perpendicular from $A$ to $B C$ is

(2, 0, - 1)

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(0, - 2, - 5)

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(3, 4, 5)

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(4, 3, 5)

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Solution

The correct option is C $(3, 4, 5)$
Let there be a point $D$ on $B C$ such that $A D$ is perpendicular to $B C$ .
Any general point on $B C$ can be written as $(1 + 4 k, 4, 6 - 2 k)$
Now dr's of $B C = (4, 0, - 2)$
dr's of $A D = (4 k, 2, 5 - 2 k)$
Since $A D$ is perpendicular to $B C, A D . B C = 0$ ,
$16 k - 10 + 4 k = 0$
i.e $k = 0.5$
Putting the value of $k$ back in $D$ , we get $D (3, 4, 5)$

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