If the distance between the plane Ax - 2y + z | vedclass

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(D) The equation of the plane containing the two lines is given by $a(x - 2) + b(y - 3) + c(z - 4) = 0$,where the normal vector $(a, b, c)$ is perpend

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$6$

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(D) The equation of the plane containing the two lines is given by $a(x - 2) + b(y - 3) + c(z - 4) = 0$,where the normal vector $(a, b, c)$ is perpendicular to the direction vectors $(2, 3, 4)$ and $(3, 4, 5)$.Solving the system $2a + 3b + 4c = 0$ and $3a + 4b + 5c = 0$,we find the cross product: $\vec{n} = (2, 3, 4) 	imes (3, 4, 5) = (15-16, 12-10, 8-9) = (-1, 2, -1)$.Thus,the equation of the plane is $-1(x - 2) + 2(y - 3) - 1(z - 4) = 0$,which simplifies to $-x + 2y - z = 0$ or $x - 2y + z = 0$.Comparing this with the given plane $Ax - 2y + z = d$,we must have $A = 1$ for the planes to be parallel.The distance between the parallel planes $x - 2y + z = 0$ and $x - 2y + z = d$ is given by $\frac{|d - 0|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \sqrt{6}$.This gives $\frac{|d|}{\sqrt{6}} = \sqrt{6}$,which implies $|d| = 6$.

If the distance between the plane $Ax - 2y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $\sqrt{6}$,then $|d|$ is

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