Let P be the plane passing through the inters | vedclass

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(C) The equation of the family of planes passing through the intersection of the given planes is $(x + 3y - z - 5) + \lambda(2x - y + z - 3) = 0$.Sinc

Vedclass · Accepted Answer

$X$ and $Y$ are on the opposite sides of $P$

Vedclass · Answer

(C) The equation of the family of planes passing through the intersection of the given planes is $(x + 3y - z - 5) + \lambda(2x - y + z - 3) = 0$.Since the plane passes through $(2, 1, -2)$,we substitute these coordinates into the equation:$(2 + 3(1) - (-2) - 5) + \lambda(2(2) - 1 + (-2) - 3) = 0$$(2 + 3 + 2 - 5) + \lambda(4 - 1 - 2 - 3) = 0$$2 + \lambda(-2) = 0 \Rightarrow \lambda = 1$.Substituting $\lambda = 1$ into the equation,we get the plane $P: 3x + 2y - 8 = 0$.Let $f(x, y, z) = 3x + 2y - 8$. We evaluate $f$ at the given points:For $X(1, -2, 4): f(1, -2, 4) = 3(1) + 2(-2) - 8 = 3 - 4 - 8 = -9$.For $Y(5, -1, 2): f(5, -1, 2) = 3(5) + 2(-1) - 8 = 15 - 2 - 8 = 5$.Since $f(X) = -9$ and $f(Y) = 5$ have opposite signs,$X$ and $Y$ lie on opposite sides of the plane $P$.

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