The distance of the line 3y - 2z - 1 = 0 = 3x | vedclass

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(C) The line is given by the intersection of two planes: $3y - 2z - 1 = 0$ and $3x - z + 4 = 0$.The direction vector $\vec{v}$ of the line is the cros

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$2\sqrt{6}$

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(C) The line is given by the intersection of two planes: $3y - 2z - 1 = 0$ and $3x - z + 4 = 0$.The direction vector $\vec{v}$ of the line is the cross product of the normals to the planes,$\vec{n_1} = (0, 3, -2)$ and $\vec{n_2} = (3, 0, -1)$.$\vec{v} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 3 & -2 \ 3 & 0 & -1 \end{vmatrix} = \hat{i}(-3) - \hat{j}(6) + \hat{k}(-9) = (-3, -6, -9)$.We can simplify the direction ratios to $(1, 2, 3)$.To find a point on the line,let $z = k$. Then $3y = 2k + 1 \Rightarrow y = \frac{2k+1}{3}$ and $3x = k - 4 \Rightarrow x = \frac{k-4}{3}$.Setting $k = 1$,we get the point $P = (-1, 1, 1)$.The line equation is $\frac{x+1}{1} = \frac{y-1}{2} = \frac{z-1}{3} = \lambda$.Any point $Q$ on the line is $(\lambda - 1, 2\lambda + 1, 3\lambda + 1)$.The vector $\vec{PQ} = (\lambda - 1 - 2, 2\lambda + 1 - (-1), 3\lambda + 1 - 6) = (\lambda - 3, 2\lambda + 2, 3\lambda - 5)$.Since $\vec{PQ}$ is perpendicular to the line direction $(1, 2, 3)$,their dot product is zero:$1(\lambda - 3) + 2(2\lambda + 2) + 3(3\lambda - 5) = 0$.$\lambda - 3 + 4\lambda + 4 + 9\lambda - 15 = 0 \Rightarrow 14\lambda - 14 = 0 \Rightarrow \lambda = 1$.The point $Q$ is $(1-1, 2+1, 3+1) = (0, 3, 4)$.The distance from $(2, -1, 6)$ to $(0, 3, 4)$ is $\sqrt{(0-2)^2 + (3 - (-1))^2 + (4-6)^2} = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}$.

The distance of the line $3y - 2z - 1 = 0 = 3x - z + 4$ from the point $(2, -1, 6)$ is:

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