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(A) Let the line be $\frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3} = \lambda$.Any point $P$ on the line is given by $(4\lambda - 2, 2\lambda + 1, 3\la

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$5$

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(A) Let the line be $\frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3} = \lambda$.Any point $P$ on the line is given by $(4\lambda - 2, 2\lambda + 1, 3\lambda - 1)$.Let $A = (1, 2, 4)$. The vector $\vec{AP} = (4\lambda - 2 - 1, 2\lambda + 1 - 2, 3\lambda - 1 - 4) = (4\lambda - 3, 2\lambda - 1, 3\lambda - 5)$.The direction vector of the line is $\vec{b} = 4\hat{i} + 2\hat{j} + 3\hat{k}$.Since $AP \perp 	ext{line}$,$\vec{AP} \cdot \vec{b} = 0$.$4(4\lambda - 3) + 2(2\lambda - 1) + 3(3\lambda - 5) = 0$.$16\lambda - 12 + 4\lambda - 2 + 9\lambda - 15 = 0$.$29\lambda - 29 = 0 \implies \lambda = 1$.Substituting $\lambda = 1$ in the coordinates of $P$,we get $P = (4(1) - 2, 2(1) + 1, 3(1) - 1) = (2, 3, 2)$.The distance of $P(2, 3, 2)$ from the plane $3x + 4y + 12z + 23 = 0$ is given by $d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$.$d = \frac{|3(2) + 4(3) + 12(2) + 23|}{\sqrt{3^2 + 4^2 + 12^2}} = \frac{|6 + 12 + 24 + 23|}{\sqrt{9 + 16 + 144}} = \frac{|65|}{\sqrt{169}} = \frac{65}{13} = 5$.

Let the foot of the perpendicular from the point $(1, 2, 4)$ on the line $\frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3}$ be $P$. Then the distance of $P$ from the plane $3x + 4y + 12z + 23 = 0$ is:

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