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(A) Let a point $P$ on the line $\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{1} = \lambda$ be $(2\lambda + 1, -\lambda - 1, \lambda)$.Since the foot

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$(2, 0, 1)$

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(A) Let a point $P$ on the line $\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{1} = \lambda$ be $(2\lambda + 1, -\lambda - 1, \lambda)$.Since the foot of the perpendicular $Q$ lies on both planes $x + y + z = 3$ and $x - y + z = 3$,we can find the intersection of these two planes.Adding the two equations: $(x + y + z) + (x - y + z) = 3 + 3 \Rightarrow 2x + 2z = 6 \Rightarrow x + z = 3$.Subtracting the two equations: $(x + y + z) - (x - y + z) = 3 - 3 \Rightarrow 2y = 0 \Rightarrow y = 0$.Since $Q$ lies on the line of intersection of the two planes,its coordinates must satisfy $y = 0$ and $z = 3 - x$.Thus,$Q$ is of the form $(x, 0, 3 - x)$.Since $PQ$ is perpendicular to the plane $x + y + z = 3$,the vector $\vec{PQ}$ must be parallel to the normal vector of the plane,$\vec{n} = (1, 1, 1)$.$\vec{PQ} = (x - (2\lambda + 1), 0 - (-\lambda - 1), 3 - x - \lambda) = (x - 2\lambda - 1, \lambda + 1, 3 - x - \lambda)$.Since $\vec{PQ} = k(1, 1, 1)$,we have $x - 2\lambda - 1 = \lambda + 1 = 3 - x - \lambda$.From $\lambda + 1 = 3 - x - \lambda$,we get $2\lambda + x = 2$.From $x - 2\lambda - 1 = \lambda + 1$,we get $x - 3\lambda = 2$.Solving $2\lambda + x = 2$ and $x - 3\lambda = 2$,we subtract the equations: $5\lambda = 0 \Rightarrow \lambda = 0$.Substituting $\lambda = 0$ into $x - 3\lambda = 2$,we get $x = 2$.Thus,the coordinates of $Q$ are $(2, 0, 3 - 2) = (2, 0, 1)$.

$A$ perpendicular is drawn from a point $P$ on the line $\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{1}$ to the plane $x + y + z = 3$ such that the foot of the perpendicular $Q$ also lies on the plane $x - y + z = 3$. Then the coordinates of $Q$ are

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