The equation of the plane in which the lines | vedclass

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(A) The equation of the plane containing the line $\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{-5}$ is given by $A(x - 5) + B(y - 7) + C(z + 3) =

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$17x - 47y - 24z + 172 = 0$

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(A) The equation of the plane containing the line $\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{-5}$ is given by $A(x - 5) + B(y - 7) + C(z + 3) = 0$ where $4A + 4B - 5C = 0$ $(i)$.Since the second line $\frac{x - 8}{7} = \frac{y - 4}{1} = \frac{z - 5}{3}$ also lies in the plane,the point $(8, 4, 5)$ must satisfy the plane equation: $A(8 - 5) + B(4 - 7) + C(5 + 3) = 0$,which simplifies to $3A - 3B + 8C = 0$ (ii).Also,the direction vector of the second line must be perpendicular to the normal vector $(A, B, C)$,so $7A + 1B + 3C = 0$ (iii).Solving equations $(i)$,(ii),and (iii) for the ratios of $A, B, C$ using the cross product of the direction vectors $(4, 4, -5)$ and $(7, 1, 3)$:Normal vector $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 4 & -5 \ 7 & 1 & 3 \end{vmatrix} = \hat{i}(12 + 5) - \hat{j}(12 + 35) + \hat{k}(4 - 28) = 17\hat{i} - 47\hat{j} - 24\hat{k}$.Thus,$A = 17, B = -47, C = -24$.The equation of the plane is $17(x - 5) - 47(y - 7) - 24(z + 3) = 0$.$17x - 85 - 47y + 329 - 24z - 72 = 0$.$17x - 47y - 24z + 172 = 0$.

The equation of the plane in which the lines $\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{-5}$ and $\frac{x - 8}{7} = \frac{y - 4}{1} = \frac{z - 5}{3}$ lie,is

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