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

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(C) The midpoint $M$ of the line segment $PQ$ is given by $M = \left(\frac{4+2}{2}, \frac{-3+3}{2}, \frac{1-5}{2}\right) = (3, 0, -2)$.Since the plane

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$28$

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(C) The midpoint $M$ of the line segment $PQ$ is given by $M = \left(\frac{4+2}{2}, \frac{-3+3}{2}, \frac{1-5}{2}\right) = (3, 0, -2)$.Since the plane bisects the line segment $PQ$ at right angles,the normal vector to the plane is parallel to the vector $\vec{PQ} = (2-4, 3-(-3), -5-1) = (-2, 6, -6)$.We can take the normal vector as $\vec{n} = (1, -3, 3)$ by dividing by $-2$.The equation of the plane passing through $M(3, 0, -2)$ with normal vector $\vec{n} = (1, -3, 3)$ is:$1(x-3) - 3(y-0) + 3(z+2) = 0$$x - 3y + 3z - 3 + 6 = 0$$x - 3y + 3z + 3 = 0$Comparing this with $ax+by+cz+d=0$,we get $a=1, b=-3, c=3, d=3$.The value of $(a^{2}+b^{2}+c^{2}+d^{2})$ is $1^{2} + (-3)^{2} + 3^{2} + 3^{2} = 1 + 9 + 9 + 9 = 28$.Since $a, b, c, d$ are integers and the plane equation is unique up to a scalar multiple,the minimum value for integer coefficients is $28$.

Let the plane $ax+by+cz+d=0$ bisect the line segment joining the points $P(4,-3,1)$ and $Q(2,3,-5)$ 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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