The distance of the point (2, 3, 4) from the | vedclass

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(D) The given line is $1 - x = \frac{y}{2} = \frac{1}{3}(1 + z)$.Rewriting the line in standard form: $\frac{x - 1}{-1} = \frac{y - 0}{2} = \frac{z -

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$\frac{3}{7}\sqrt{35}$

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(D) The given line is $1 - x = \frac{y}{2} = \frac{1}{3}(1 + z)$.Rewriting the line in standard form: $\frac{x - 1}{-1} = \frac{y - 0}{2} = \frac{z - (-1)}{3}$.Let the point $P = (2, 3, 4)$ and the line pass through $A = (1, 0, -1)$ with direction vector $\vec{v} = (-1, 2, 3)$.The vector $\vec{AP} = (2-1, 3-0, 4-(-1)) = (1, 3, 5)$.The distance $d$ of point $P$ from the line is given by $d = \frac{|\vec{AP} 	imes \vec{v}|}{|\vec{v}|}$.$\vec{AP} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 3 & 5 \ -1 & 2 & 3 \end{vmatrix} = \hat{i}(9 - 10) - \hat{j}(3 + 5) + \hat{k}(2 + 3) = -\hat{i} - 8\hat{j} + 5\hat{k}$.$|\vec{AP} 	imes \vec{v}| = \sqrt{(-1)^2 + (-8)^2 + 5^2} = \sqrt{1 + 64 + 25} = \sqrt{90} = 3\sqrt{10}$.$|\vec{v}| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.$d = \frac{3\sqrt{10}}{\sqrt{14}} = \frac{3\sqrt{5}}{\sqrt{7}} = \frac{3\sqrt{35}}{7} = \frac{3}{7}\sqrt{35}$.

The distance of the point $(2, 3, 4)$ from the line $1 - x = \frac{y}{2} = \frac{1}{3}(1 + z)$ is

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