If (alpha, beta, gamma) is the foot of the pe | vedclass

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(B) The line passes through $A(2, -1, 1)$ and $B(1, 1, -2)$. The direction vector of the line is $\vec{v} = (1-2, 1-(-1), -2-1) = (-1, 2, -3)$.The equ

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$-\frac{1}{7}$

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(B) The line passes through $A(2, -1, 1)$ and $B(1, 1, -2)$. The direction vector of the line is $\vec{v} = (1-2, 1-(-1), -2-1) = (-1, 2, -3)$.The equation of the line is $\frac{x-2}{-1} = \frac{y+1}{2} = \frac{z-1}{-3} = k$.Any point on the line is given by $(\alpha, \beta, \gamma) = (2-k, -1+2k, 1-3k)$.The vector $\vec{PQ} = (\alpha - (-1), \beta - 2, \gamma - (-1)) = (3-k, -3+2k, 2-3k)$ must be perpendicular to the line direction $\vec{v} = (-1, 2, -3)$.Thus, $\vec{PQ} \cdot \vec{v} = 0 \implies -1(3-k) + 2(-3+2k) - 3(2-3k) = 0$.$-3 + k - 6 + 4k - 6 + 9k = 0 \implies 14k - 15 = 0 \implies k = \frac{15}{14}$.Substituting $k$ back: $\alpha = 2 - \frac{15}{14} = \frac{13}{14}$, $\beta = -1 + 2(\frac{15}{14}) = \frac{16}{14}$, $\gamma = 1 - 3(\frac{15}{14}) = -\frac{31}{14}$.$\alpha + \beta + \gamma = \frac{13+16-31}{14} = -\frac{2}{14} = -\frac{1}{7}$.

If $(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from a point $P(-1, 2, -1)$ to the line joining the points $A(2, -1, 1)$ and $B(1, 1, -2)$, then $\alpha + \beta + \gamma =$

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