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

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(A) 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

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$\frac{-5}{6}$

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(A) 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 determinant of the matrix formed by their direction ratios and the difference of their points is zero: $\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{array}ight| = 0$ Given the lines: Line $1$: $(x_1, y_1, z_1) = (1, 2, 1)$ and $(a_1, b_1, c_1) = (2, 3, 4)$ Line $2$: $(x_2, y_2, z_2) = (3, k, 0)$ and $(a_2, b_2, c_2) = (-1, 2, 1)$ Substituting these values into the determinant: $\left|\begin{array}{ccc} 3-1 & k-2 & 0-1 \ 2 & 3 & 4 \ -1 & 2 & 1 \end{array}ight| = 0$ $\left|\begin{array}{ccc} 2 & k-2 & -1 \ 2 & 3 & 4 \ -1 & 2 & 1 \end{array}ight| = 0$ Expanding along the first row: $2(3(1) - 4(2)) - (k-2)(2(1) - 4(-1)) - 1(2(2) - 3(-1)) = 0$ $2(3 - 8) - (k-2)(2 + 4) - 1(4 + 3) = 0$ $2(-5) - (k-2)(6) - 1(7) = 0$ $-10 - 6k + 12 - 7 = 0$ $-6k - 5 = 0$ $-6k = 5$ $k = \frac{-5}{6}$

If the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-1}{4}$ and $\frac{x-3}{-1}=\frac{y-k}{2}=\frac{z}{1}$ intersect,then $k$ is equal to

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