Let the plane ax+by+cz+d=0 bisect the line jo | vedclass

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(B) Step-$1$: Find the midpoint $M$ of the line segment joining $P(4,-3,1)$ and $Q(2,3,-5)$.$M = \left(\frac{4+2}{2}, \frac{-3+3}{2}, \frac{1-5}{2}\ri

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$28$

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(B) Step-$1$: Find the midpoint $M$ of the line segment joining $P(4,-3,1)$ and $Q(2,3,-5)$.$M = \left(\frac{4+2}{2}, \frac{-3+3}{2}, \frac{1-5}{2}\right) = (3, 0, -2)$.Step-$2$: The normal vector $\vec{n}$ to the plane is the vector $\vec{PQ} = (2-4, 3-(-3), -5-1) = (-2, 6, -6)$.We can simplify the normal vector by dividing by $-2$ to get $\vec{n}' = (1, -3, 3)$.Thus,the equation of the plane is $1(x-3) - 3(y-0) + 3(z+2) = 0$.$x - 3y + 3z - 3 + 6 = 0 \Rightarrow x - 3y + 3z + 3 = 0$.Here,$a=1, b=-3, c=3, d=3$.Step-$3$: Calculate $a^2+b^2+c^2+d^2 = (1)^2 + (-3)^2 + (3)^2 + (3)^2 = 1 + 9 + 9 + 9 = 28$.

Let the plane $ax+by+cz+d=0$ bisect the line joining the points $(4,-3,1)$ and $(2,3,-5)$ at right angles. If $a, b, c, d$ are integers,then the minimum value of $(a^2+b^2+c^2+d^2)$ is

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