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

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(A) The given lines are $L_1: \frac{x - 1}{3} = \frac{y - 2}{-1} = \frac{z - \lambda}{2}$ and $L_2: \frac{x + 1}{-2} = \frac{y}{3\lambda} = \frac{z -

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$\frac{182}{36}$

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(A) The given lines are $L_1: \frac{x - 1}{3} = \frac{y - 2}{-1} = \frac{z - \lambda}{2}$ and $L_2: \frac{x + 1}{-2} = \frac{y}{3\lambda} = \frac{z - 7/2}{1/2}$.For two lines to be coplanar,the determinant of the vector connecting a point on each line and the direction vectors of the lines must be zero:$\left|\begin{array}{ccc} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{array}ight| = 0$Substituting the points $(1, 2, \lambda)$ and $(-1, 0, 7/2)$ and direction vectors $(3, -1, 2)$ and $(-2, 3\lambda, 1/2)$:$\left|\begin{array}{ccc} -1 - 1 & 0 - 2 & 7/2 - \lambda \ 3 & -1 & 2 \ -2 & 3\lambda & 1/2 \end{array}ight| = 0$$\left|\begin{array}{ccc} -2 & -2 & (7-2\lambda)/2 \ 3 & -1 & 2 \ -2 & 3\lambda & 1/2 \end{array}ight| = 0$Expanding the determinant:$-2(-1/2 - 6\lambda) + 2(3/2 + 4) + \frac{7-2\lambda}{2}(9\lambda - 2) = 0$$1 + 12\lambda + 11 + \frac{63\lambda - 14 - 18\lambda^2 + 4\lambda}{2} = 0$$12 + 12\lambda + \frac{-18\lambda^2 + 67\lambda - 14}{2} = 0$$24 + 24\lambda - 18\lambda^2 + 67\lambda - 14 = 0$$-18\lambda^2 + 91\lambda + 10 = 0 \Rightarrow 18\lambda^2 - 91\lambda - 10 = 0$Sum of values of $\lambda = -(-91)/18 = 91/18 = 182/36$.

If lines $\frac{x - 1}{3} = \frac{y - 2}{-1} = \frac{z - \lambda}{2}$ and $\frac{x + 1}{-2} = \frac{y}{3\lambda} = \frac{2z - 7}{1}$ are coplanar,then the sum of the value$(s)$ of $\lambda$ is:

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