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(C) Let $N$ be the foot of the perpendicular from the point $P(0,2,3)$ on the given line. Let $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=r$. Then any

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$(2,3,-1)$

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(C) Let $N$ be the foot of the perpendicular from the point $P(0,2,3)$ on the given line. Let $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=r$. Then any point on the line is given by $(5r-3, 2r+1, 3r-4)$. If this point is $N$, the direction ratios of the vector $\vec{NP}$ are $(5r-3-0, 2r+1-2, 3r-4-3)$, which simplifies to $(5r-3, 2r-1, 3r-7)$. Since $\vec{NP}$ is perpendicular to the line with direction ratios $(5, 2, 3)$, their dot product must be zero: $5(5r-3) + 2(2r-1) + 3(3r-7) = 0$. $25r - 15 + 4r - 2 + 9r - 21 = 0$. $38r - 38 = 0$, which gives $r = 1$. Substituting $r = 1$ into the point coordinates $(5r-3, 2r+1, 3r-4)$, we get $(5(1)-3, 2(1)+1, 3(1)-4) = (2, 3, -1)$.

The foot of the perpendicular from $(0,2,3)$ to the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$ is

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