If the equation of the plane bisecting the li | vedclass

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(A) The mid-point $R$ of the line segment joining $P(3,2,4)$ and $Q(-1,0,-2)$ is given by: $R = \left( \frac{3-1}{2}, \frac{2+0}{2}, \frac{4-2}{2} \ri

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(A) The mid-point $R$ of the line segment joining $P(3,2,4)$ and $Q(-1,0,-2)$ is given by: $R = \left( \frac{3-1}{2}, \frac{2+0}{2}, \frac{4-2}{2} \right) = (1, 1, 1)$. The direction ratios of the line segment $PQ$ are $(3 - (-1), 2 - 0, 4 - (-2)) = (4, 2, 6)$. Since the plane is perpendicular to $PQ$,the normal vector to the plane is $\vec{n} = \langle 4, 2, 6 \rangle$. Thus,the equation of the plane is $4x + 2y + 6z + d = 0$. Since the plane passes through the mid-point $R(1, 1, 1)$,we have: $4(1) + 2(1) + 6(1) + d = 0 \implies 4 + 2 + 6 + d = 0 \implies d = -12$. Comparing $4x + 2y + 6z - 12 = 0$ with $ax + by + cz + d = 0$,we get $a=4, b=2, c=6, d=-12$. Therefore,$ac + bd = (4)(6) + (2)(-12) = 24 - 24 = 0$.

If the equation of the plane bisecting the line segment joining the points $P(3,2,4)$ and $Q(-1,0,-2)$ and perpendicular to $PQ$ is $ax+by+cz+d=0$,then find the value of $ac+bd$.

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