If (a, b, c) is the image of the point (1, 2, | vedclass

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

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$2$

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(B) Let the given line be $\frac{x+1}{2} = \frac{y-3}{-2} = \frac{z}{-1} = \lambda$. Any point $R$ on the line is given by $(2\lambda-1, -2\lambda+3, -\lambda)$. Let $P = (1, 2, -3)$ and $Q = (a, b, c)$ be the image of $P$ in the line. The vector $\vec{PQ} = (a-1, b-2, c+3)$ must be perpendicular to the line,whose direction ratios are $(2, -2, -1)$. Thus,$2(a-1) - 2(b-2) - 1(c+3) = 0 \implies 2a - 2b - c = 1$. Also,the midpoint $R$ of $PQ$ lies on the line: $R = \left(\frac{a+1}{2}, \frac{b+2}{2}, \frac{c-3}{2}\right)$. Since $R$ lies on the line,we have: $\frac{\frac{a+1}{2} + 1}{2} = \frac{\frac{b+2}{2} - 3}{-2} = \frac{\frac{c-3}{2}}{-1} = \lambda$. Solving for $a, b, c$ in terms of $\lambda$: $a = 4\lambda - 3, b = -4\lambda + 4, c = -2\lambda + 3$. Substituting these into the perpendicularity condition $2a - 2b - c = 1$: $2(4\lambda - 3) - 2(-4\lambda + 4) - (-2\lambda + 3) = 1 \implies 8\lambda - 6 + 8\lambda - 8 + 2\lambda - 3 = 1 \implies 18\lambda = 18 \implies \lambda = 1$. Thus,$a = 4(1) - 3 = 1$,$b = -4(1) + 4 = 0$,$c = -2(1) + 3 = 1$. Therefore,$a+b+c = 1 + 0 + 1 = 2$.

If $(a, b, c)$ is the image of the point $(1, 2, -3)$ in the line $\frac{x+1}{2} = \frac{y-3}{-2} = \frac{z}{-1}$,then $a+b+c$ is equal to

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