If the lines frac{x-1}{k} = frac{y-2}{2} = fr | vedclass

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(A) The given lines are $\frac{x-1}{k} = \frac{y-2}{2} = \frac{z-3}{3}$ and $\frac{x-2}{3} = \frac{y-3}{k} = \frac{z-1}{2}$.Two lines $\frac{x-x_1}{a_

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$-5$

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(A) The given lines are $\frac{x-1}{k} = \frac{y-2}{2} = \frac{z-3}{3}$ and $\frac{x-2}{3} = \frac{y-3}{k} = \frac{z-1}{2}$.Two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ intersect if and only if the shortest distance between them is $0$,which implies the determinant condition:$\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix} = 0$.Here,$(x_1, y_1, z_1) = (1, 2, 3)$,$(a_1, b_1, c_1) = (k, 2, 3)$ and $(x_2, y_2, z_2) = (2, 3, 1)$,$(a_2, b_2, c_2) = (3, k, 2)$.Substituting these values into the determinant:$\begin{vmatrix} 2-1 & 3-2 & 1-3 \ k & 2 & 3 \ 3 & k & 2 \end{vmatrix} = 0 \implies \begin{vmatrix} 1 & 1 & -2 \ k & 2 & 3 \ 3 & k & 2 \end{vmatrix} = 0$.Expanding the determinant:$1(4-3k) - 1(2k-9) - 2(k^2-6) = 0$$4 - 3k - 2k + 9 - 2k^2 + 12 = 0$$-2k^2 - 5k + 25 = 0$$2k^2 + 5k - 25 = 0$.Factoring the quadratic equation:$2k^2 + 10k - 5k - 25 = 0$$2k(k+5) - 5(k+5) = 0$$(2k-5)(k+5) = 0$.Thus,$k = \frac{5}{2}$ or $k = -5$.Since the question asks for the integer value of $k$,the correct value is $-5$.

If the lines $\frac{x-1}{k} = \frac{y-2}{2} = \frac{z-3}{3}$ and $\frac{x-2}{3} = \frac{y-3}{k} = \frac{z-1}{2}$ intersect each other,then the integer value of $k$ is:

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