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(A) The direction ratios of the line are given by the cross product of the normals to the planes $x + y - z = 0$ and $x - 2y + 3z - 5 = 0$.$\vec{v} =

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$\sqrt{\frac{21}{2}}$

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(A) The direction ratios of the line are given by the cross product of the normals to the planes $x + y - z = 0$ and $x - 2y + 3z - 5 = 0$.$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & -1 \ 1 & -2 & 3 \end{vmatrix} = \hat{i}(3 - 2) - \hat{j}(3 + 1) + \hat{k}(-2 - 1) = \hat{i} - 4\hat{j} - 3\hat{k}$.The equation of the line passing through $(1, 2, 4)$ with direction vector $\vec{v} = (1, -4, -3)$ is $\vec{r} = (1, 2, 4) + \lambda(1, -4, -3)$.Let $P$ be a point on the line: $P = (1 + \lambda, 2 - 4\lambda, 4 - 3\lambda)$.Let $A = (1, -2, 5)$. The vector $\vec{AP} = P - A = (\lambda, 4 - 4\lambda, -1 - 3\lambda)$.Since $\vec{AP}$ is perpendicular to the line direction $\vec{v} = (1, -4, -3)$,their dot product is zero:$\vec{AP} \cdot \vec{v} = 1(\lambda) - 4(4 - 4\lambda) - 3(-1 - 3\lambda) = 0$.$\lambda - 16 + 16\lambda + 3 + 9\lambda = 0 \implies 26\lambda - 13 = 0 \implies \lambda = \frac{1}{2}$.Substituting $\lambda = \frac{1}{2}$ into $P$,we get $P = (1 + \frac{1}{2}, 2 - 2, 4 - \frac{3}{2}) = (\frac{3}{2}, 0, \frac{5}{2})$.The length of the perpendicular is the distance $AP = \sqrt{(\frac{3}{2} - 1)^2 + (0 - (-2))^2 + (\frac{5}{2} - 5)^2} = \sqrt{(\frac{1}{2})^2 + (2)^2 + (-\frac{5}{2})^2} = \sqrt{\frac{1}{4} + 4 + \frac{25}{4}} = \sqrt{\frac{26}{4} + \frac{16}{4}} = \sqrt{\frac{42}{4}} = \sqrt{\frac{21}{2}}$.

The length of the perpendicular from the point $(1, -2, 5)$ to the line passing through $(1, 2, 4)$ and parallel to the line $x + y - z = 0 = x - 2y + 3z - 5$ is:

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