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(B) Let the point $P = (1,0,-2)$ and the foot of the perpendicular $F = (2,0,-1)$.The direction ratios of the normal to the plane are given by the vec

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$8$

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(B) Let the point $P = (1,0,-2)$ and the foot of the perpendicular $F = (2,0,-1)$.The direction ratios of the normal to the plane are given by the vector $\vec{PF} = (2-1, 0-0, -1-(-2)) = (1, 0, 1)$.Since the equation of the plane is $ax+by+cz=2$, the normal vector is $\vec{n} = (a, b, c)$.Thus, $(a, b, c) = \lambda(1, 0, 1) = (\lambda, 0, \lambda)$ for some constant $\lambda$.The equation of the plane becomes $\lambda x + 0y + \lambda z = 2$, or $\lambda(x+z) = 2$.Since the point $F(2,0,-1)$ lies on the plane, we substitute its coordinates into the equation:$\lambda(2 + (-1)) = 2 \implies \lambda(1) = 2 \implies \lambda = 2$.Therefore, $a = \lambda = 2$, $b = 0$, and $c = \lambda = 2$.Finally, $a^2+b^2+c^2 = 2^2 + 0^2 + 2^2 = 4 + 0 + 4 = 8$.

If the foot of the perpendicular drawn from the point $(1,0,-2)$ to the plane $\pi$ is $(2,0,-1)$ and the equation of the plane $\pi$ is $ax+by+cz=2$, then $a^2+b^2+c^2=$

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