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

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(D) The equation of the plane containing the two given lines is given by the determinant form:$\begin{vmatrix} x-1 & y-2 & z-3 \ 2 & 3 & 4 \ 3 & 4 &

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$6$

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(D) The equation of the plane containing the two given lines is given by the determinant form:$\begin{vmatrix} x-1 & y-2 & z-3 \ 2 & 3 & 4 \ 3 & 4 & 5 \end{vmatrix} = 0$Expanding along the first row:$(x-1)(15-16) - (y-2)(10-12) + (z-3)(8-9) = 0$$-(x-1) + 2(y-2) - (z-3) = 0$$-x + 1 + 2y - 4 - z + 3 = 0$$-x + 2y - z = 0 \implies x - 2y + z = 0$ ... $(i)$Given the plane equation $Ax - 2y + z = d$ ... $(ii)$Since the planes are parallel,the coefficients of $x, y, z$ must be proportional. Thus,$A = 1$.The distance $D$ between two parallel planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$ is given by $D = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$.Here,$D = \sqrt{6}$,$a=1, b=-2, c=1$,$d_1 = 0$,and $d_2 = d$.$\sqrt{6} = \frac{|0 - d|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|d|}{\sqrt{6}}$$|d| = \sqrt{6} 	imes \sqrt{6} = 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 $\sqrt{6}$ units,then $|d|$ is

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