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

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(C) The direction vectors of the two lines are $\vec{v_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{v_2} = 3\hat{i} + 4\hat{j} + 5\hat{k}$.The norma

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$6$

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(C) The direction vectors of the two lines are $\vec{v_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{v_2} = 3\hat{i} + 4\hat{j} + 5\hat{k}$.The normal vector $\vec{n}$ to the plane containing these lines is $\vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9) = -\hat{i} + 2\hat{j} - \hat{k}$.Since the plane passes through the point $(1, 2, 3)$,its equation is $-1(x-1) + 2(y-2) - 1(z-3) = 0$,which simplifies to $-x + 2y - z = 0$ or $x - 2y + z = 0$.The given plane is $Ax - 2y + z = d$. Comparing the coefficients,we find $A = 1$.The distance between the parallel planes $x - 2y + z = 0$ and $x - 2y + z = d$ is given by $\frac{|d - 0|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \sqrt{6}$.$\frac{|d|}{\sqrt{6}} = \sqrt{6} \implies |d| = 6$.

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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