If the lines frac{x-k}{2}=frac{y+1}{3}=frac{z | vedclass

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(C) For two lines to intersect,the shortest distance between them must be $0$. The condition for intersection of two lines $\frac{x-x_1}{a_1} = \frac{

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(C) For two lines to intersect,the shortest distance between them must be $0$. The condition for intersection of two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ is given by the determinant: $\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{array}ight| = 0$. Substituting the given values: $(x_1, y_1, z_1) = (k, -1, 1)$,$(a_1, b_1, c_1) = (2, 3, 4)$ and $(x_2, y_2, z_2) = (3, \frac{9}{2}, 0)$,$(a_2, b_2, c_2) = (1, 2, 1)$. The determinant becomes: $\left|\begin{array}{ccc} 3-k & \frac{9}{2}-(-1) & 0-1 \ 2 & 3 & 4 \ 1 & 2 & 1 \end{array}ight| = 0$. $\Rightarrow \left|\begin{array}{ccc} 3-k & \frac{11}{2} & -1 \ 2 & 3 & 4 \ 1 & 2 & 1 \end{array}ight| = 0$. Expanding the determinant: $(3-k)(3-8) - \frac{11}{2}(2-4) - 1(4-3) = 0$. $\Rightarrow (3-k)(-5) - \frac{11}{2}(-2) - 1(1) = 0$. $\Rightarrow -15 + 5k + 11 - 1 = 0$. $\Rightarrow 5k - 5 = 0$. $\Rightarrow 5k = 5$. $\Rightarrow k = 1$.

If the lines $\frac{x-k}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-\frac{9}{2}}{2}=\frac{z}{1}$ intersect,then the value of $k$ is

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