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(C) Let the point $P = (-2, -1, 3)$ and the foot of the perpendicular $F = (1, 0, -2)$.The normal vector $\vec{n}$ to the plane $\pi$ is parallel to t

Vedclass · Accepted Answer

$13$

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(C) Let the point $P = (-2, -1, 3)$ and the foot of the perpendicular $F = (1, 0, -2)$.The normal vector $\vec{n}$ to the plane $\pi$ is parallel to the vector $\vec{PF} = (1 - (-2), 0 - (-1), -2 - 3) = (3, 1, -5)$.Thus,the equation of the plane passing through $F(1, 0, -2)$ with normal vector $\vec{n} = (3, 1, -5)$ is:$3(x - 1) + 1(y - 0) - 5(z + 2) = 0$$3x - 3 + y - 5z - 10 = 0$$3x + y - 5z = 13$Dividing by $13$,we get the intercept form:$\frac{x}{13/3} + \frac{y}{13} + \frac{z}{-13/5} = 1$Comparing with $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,we have $a = \frac{13}{3}$,$b = 13$,and $c = -\frac{13}{5}$.Now,calculate $3a + b + 5c$:$3(\frac{13}{3}) + 13 + 5(-\frac{13}{5}) = 13 + 13 - 13 = 13$.

The foot of the perpendicular drawn from the point $(-2,-1,3)$ to a plane $\pi$ is $(1,0,-2)$. If $a, b, c$ are the intercepts made by the plane $\pi$ on $X, Y, Z$-axes respectively,then $3a+b+5c=$

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