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

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(C) Let the lines be $L_1: \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} = \lambda$ and $L_2: \frac{x - 2}{1} = \frac{y + m}{2} = \frac{z - 2}{1

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$-1$

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(C) Let the lines be $L_1: \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} = \lambda$ and $L_2: \frac{x - 2}{1} = \frac{y + m}{2} = \frac{z - 2}{1} = \mu$.Any point on $L_1$ is $P(2\lambda + 1, 3\lambda - 1, 4\lambda + 1)$ and any point on $L_2$ is $Q(\mu + 2, 2\mu - m, \mu + 2)$.For intersection, $P = Q$, so:$2\lambda + 1 = \mu + 2 \implies 2\lambda - \mu = 1$ $(i)$$3\lambda - 1 = 2\mu - m \implies 3\lambda - 2\mu = -m - 1$ (ii)$4\lambda + 1 = \mu + 2 \implies 4\lambda - \mu = 1$ (iii)Subtracting $(i)$ from (iii): $(4\lambda - \mu) - (2\lambda - \mu) = 1 - 1 \implies 2\lambda = 0 \implies \lambda = 0$.Substituting $\lambda = 0$ in $(i)$: $2(0) - \mu = 1 \implies \mu = -1$.Substituting $\lambda = 0$ and $\mu = -1$ in (ii): $3(0) - 2(-1) = -m - 1 \implies 2 = -m - 1 \implies m = -3$. Wait, re-evaluating: $2 = -m - 1 \implies m = -3$. Checking options, let's re-solve: $3(0) - 2(-1) = -m - 1 \implies 2 = -m - 1 \implies m = -3$. Since $-3$ is not in options, let's re-check the intersection condition. The lines intersect if the determinant of the vector difference and direction vectors is zero: $\begin{vmatrix} 2-1 & -m-(-1) & 2-1 \ 2 & 3 & 4 \ 1 & 2 & 1 \end{vmatrix} = 0 \implies \begin{vmatrix} 1 & 1-m & 1 \ 2 & 3 & 4 \ 1 & 2 & 1 \end{vmatrix} = 0$.$1(3-8) - (1-m)(2-4) + 1(4-3) = 0 \implies -5 + 2(1-m) + 1 = 0 \implies -4 + 2 - 2m = 0 \implies -2 - 2m = 0 \implies m = -1$.

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

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