If the lines frac{x - a + d}{alpha - delta} = | vedclass

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Question

(C) The condition for two lines to be coplanar is that the determinant of the coordinates of a point on each line and their direction ratios must be z

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$x - 2y + z = 0$

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(C) The condition for two lines to be coplanar is that the determinant of the coordinates of a point on each line and their direction ratios must be zero.The equation of the plane containing the lines is given by the determinant:$\begin{vmatrix} x - a + d & y - a & z - a - d \ \alpha - \delta & \alpha & \alpha + \delta \ \beta - \gamma & \beta & \beta + \gamma \end{vmatrix} = 0$Perform the column operation $C_1 	o C_1 + C_3$:$\begin{vmatrix} x + z - 2a & y - a & z - a - d \ 2\alpha & \alpha & \alpha + \delta \ 2\beta & \beta & \beta + \gamma \end{vmatrix} = 0$Now,perform $C_1 	o C_1 - 2C_2$:$\begin{vmatrix} x + z - 2y & y - a & z - a - d \ 0 & \alpha & \alpha + \delta \ 0 & \beta & \beta + \gamma \end{vmatrix} = 0$Expanding along the first column:$(x + z - 2y) [\alpha(\beta + \gamma) - \beta(\alpha + \delta)] = 0$Assuming the direction vectors are not parallel,the term in the square bracket is non-zero.Therefore,$x + z - 2y = 0$,which is $x - 2y + z = 0$.

If the lines $\frac{x - a + d}{\alpha - \delta} = \frac{y - a}{\alpha} = \frac{z - a - d}{\alpha + \delta}$ and $\frac{x - b + c}{\beta - \gamma} = \frac{y - b}{\beta} = \frac{z - b - c}{\beta + \gamma}$ are coplanar,then the equation of the plane containing them is .........

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