Find the distance of the point -hat{i} - 5hat | vedclass

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(D) Let the coordinates of any point on the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12} = r$ be $(3r + 2, 4r - 1, 12r + 2)$.Since this

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$13$

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(D) Let the coordinates of any point on the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12} = r$ be $(3r + 2, 4r - 1, 12r + 2)$.Since this point lies on the plane $x - y + z = 5$,we substitute the coordinates into the plane equation:$(3r + 2) - (4r - 1) + (12r + 2) = 5$$3r - 4r + 12r + 2 + 1 + 2 = 5$$11r + 5 = 5$$11r = 0 \implies r = 0$.Substituting $r = 0$ into the coordinates,the point of intersection is $(2, -1, 2)$.The distance between the point $(-1, -5, -10)$ and the intersection point $(2, -1, 2)$ is given by the distance formula:$d = \sqrt{(2 - (-1))^2 + (-1 - (-5))^2 + (2 - (-10))^2}$$d = \sqrt{(3)^2 + (4)^2 + (12)^2}$$d = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$.

Find the distance of the point $-\hat{i} - 5\hat{j} - 10\hat{k}$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 5$.

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