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(B) Let the line be $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3} = k$.Any point $P$ on the line is given by $P(5k-3, 2k+1, 3k-4)$.The direction rati

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$101$

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(B) Let the line be $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3} = k$.Any point $P$ on the line is given by $P(5k-3, 2k+1, 3k-4)$.The direction ratios of the line $AP$,where $A$ is $(1, 2, 3)$,are $(5k-3-1, 2k+1-2, 3k-4-3) = (5k-4, 2k-1, 3k-7)$.The direction ratios of the given line are $(5, 2, 3)$.Since $AP$ is perpendicular to the line,the dot product of their direction ratios must be zero:$5(5k-4) + 2(2k-1) + 3(3k-7) = 0$$25k - 20 + 4k - 2 + 9k - 21 = 0$$38k - 43 = 0 \implies k = \frac{43}{38}$.The coordinates of the foot $P$ are $(\alpha, \beta, \gamma) = (5k-3, 2k+1, 3k-4)$.Then $\alpha + \beta + \gamma = (5k-3) + (2k+1) + (3k-4) = 10k - 6$.Substituting $k = \frac{43}{38}$:$\alpha + \beta + \gamma = 10\left(\frac{43}{38}ight) - 6 = \frac{430 - 228}{38} = \frac{202}{38} = \frac{101}{19}$.Therefore,$19(\alpha + \beta + \gamma) = 19 	imes \frac{101}{19} = 101$.

Let $(\alpha, \beta, \gamma)$ be the foot of the perpendicular from the point $(1, 2, 3)$ on the line $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$. Then $19(\alpha + \beta + \gamma)$ is equal to:

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