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Collinear Vectors

If the line ...

Question

If the line $\frac{x - 3}{2} = \frac{y - 4}{3} = \frac{z - 5}{4}$ lies in the plane $4 x + 4 y - k z - d = 0$ , then the value of $k$ and $d$ , are-

A

3, 5

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B

5, 3

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C

2, 5

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D

5, 2

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Solution

The correct option is B $5, 3$
If the line lies in plane dot (.) product must be zero and the point must satisfy the plane equation
$\begin{matrix} 8 + 12 - 4 x = 0 K = 5 12 + 16 - 25 - d = 0 d = 3 \end{matrix}$

Hence, this is the answer.

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