The lines frac{x-2}{1}=frac{y-3}{1}=frac{z-4} | vedclass

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(A) The given lines are $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-K}$ and $\frac{x-1}{K}=\frac{y-4}{2}=\frac{z-5}{1}$.For two lines $\frac{x-x_1}{a_1}=

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$K=0$

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(A) The given lines are $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-K}$ and $\frac{x-1}{K}=\frac{y-4}{2}=\frac{z-5}{1}$.For two lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ to be coplanar,the condition is $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix} = 0$.Here,$(x_1, y_1, z_1) = (2, 3, 4)$ and $(x_2, y_2, z_2) = (1, 4, 5)$.Also,$(a_1, b_1, c_1) = (1, 1, -K)$ and $(a_2, b_2, c_2) = (K, 2, 1)$.Substituting these values into the determinant:$\begin{vmatrix} 1-2 & 4-3 & 5-4 \ 1 & 1 & -K \ K & 2 & 1 \end{vmatrix} = 0$$\begin{vmatrix} -1 & 1 & 1 \ 1 & 1 & -K \ K & 2 & 1 \end{vmatrix} = 0$Expanding the determinant:$-1(1 - (-2K)) - 1(1 - (-K^2)) + 1(2 - K) = 0$$-1(1 + 2K) - 1(1 + K^2) + 2 - K = 0$$-1 - 2K - 1 - K^2 + 2 - K = 0$$-K^2 - 3K = 0$$K(K + 3) = 0$Thus,$K = 0$ or $K = -3$. Since $K = -3$ is not in the options,the correct value is $K = 0$.

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 if

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