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

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(B) The lines are given by $\frac{x - 3}{2} = \frac{y + 1}{-3} = \frac{z + a}{p}$ and $\frac{x + 2}{2} = \frac{y - 4}{4} = \frac{z + 5}{2}$.Since the

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$-\frac{2}{7}$

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(B) The lines are given by $\frac{x - 3}{2} = \frac{y + 1}{-3} = \frac{z + a}{p}$ and $\frac{x + 2}{2} = \frac{y - 4}{4} = \frac{z + 5}{2}$.Since the lines are perpendicular,the dot product of their direction vectors $(2, -3, p)$ and $(2, 4, 2)$ must be zero:$(2)(2) + (-3)(4) + (p)(2) = 0$$4 - 12 + 2p = 0 \Rightarrow 2p = 8 \Rightarrow p = 4$.Since the lines are coplanar,the determinant of the vector connecting points $(3, -1, -a)$ and $(-2, 4, -5)$ and the two direction vectors must be zero:$\left|\begin{array}{ccc} 3 - (-2) & -1 - 4 & -a - (-5) \ 2 & -3 & p \ 2 & 4 & 2 \end{array}ight| = 0$$\left|\begin{array}{ccc} 5 & -5 & 5 - a \ 2 & -3 & 4 \ 2 & 4 & 2 \end{array}ight| = 0$$5(-6 - 16) + 5(4 - 8) + (5 - a)(8 + 6) = 0$$5(-22) + 5(-4) + 14(5 - a) = 0$$-110 - 20 + 70 - 14a = 0$$-60 - 14a = 0 \Rightarrow 14a = -60 \Rightarrow a = -\frac{30}{7}$.Thus,$a + p = -\frac{30}{7} + 4 = \frac{-30 + 28}{7} = -\frac{2}{7}$.

If lines $\frac{x - 3}{2} = \frac{y + 1}{-3} = \frac{z + a}{p}$ and $\frac{x + 2}{2} = \frac{y - 4}{4} = \frac{z + 5}{2}$ are perpendicular coplanar lines,then the value of $a + p$ is

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