If the lines x-2/1 = y-3/1 = z-4/k and x-1/k = y-4/2 = z-5/1 are coplanar, then k can have: - Mathematics MCQ JEE Main 2026 | ExamTest.live

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If the lines \frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{k} and \frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1} are coplanar, then k can have:

Solution & Step-by-step Explanation

For two lines \frac{x - x _{1}}{l _{1}} = \frac{y - y _{1}}{m _{1}} = \frac{z - z _{1}}{n _{1}} and \frac{x - x _{2}}{l _{2}} = \frac{y - y _{2}}{m _{2}} = \frac{z - z _{2}}{n _{2}} to be coplanar: x_{2} - x_{1} l_{1} l_{2} y_{2} - y_{1} m_{1} m_{2} z_{2} - z_{1} n_{1} n_{2} = 0 1 - 2 1 k 4 - 3 12 5 - 4 k 1 = - 1 1 k 112 1 k 1 = 0 - 1 (1 - 2 k) - 1 (1 - k^{2}) + 1 (2 - k) = 0 - 1 + 2 k - 1 + k^{2} + 2 - k = 0 k^{2} + k = 0 \Rightarrow k (k + 1) = 0 \Rightarrow k = 0, - 1 . k can have exactly two values.

Try it yourself before checking the explanation above.

If the lines \frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{k} and \frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1} are coplanar, then k can have:

exactly one value

exactly two values

exactly three values

any value

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