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(C) The equation of the family of planes passing through the intersection of the given planes is given by $(x + y + 4z - 16) + \lambda(-x + y + z - 6)

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$(4, 2, 2)$

Vedclass · Answer

(C) The equation of the family of planes passing through the intersection of the given planes is given by $(x + y + 4z - 16) + \lambda(-x + y + z - 6) = 0$.Since the plane $P$ passes through the point $(1, 2, 3)$,we substitute $x = 1, y = 2, z = 3$ into the equation:$(1 + 2 + 4(3) - 16) + \lambda(-1 + 2 + 3 - 6) = 0$$(1 + 2 + 12 - 16) + \lambda(-2) = 0$$-1 - 2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$.Substituting $\lambda = -\frac{1}{2}$ back into the family equation:$(x + y + 4z - 16) - \frac{1}{2}(-x + y + z - 6) = 0$$2(x + y + 4z - 16) - (-x + y + z - 6) = 0$$2x + 2y + 8z - 32 + x - y - z + 6 = 0$$3x + y + 7z - 26 = 0$.Now,we check which point does not satisfy the equation $3x + y + 7z = 26$:For $(3, 3, 2)$: $3(3) + 3 + 7(2) = 9 + 3 + 14 = 26$ (Lies on $P$).For $(6, -6, 2)$: $3(6) - 6 + 7(2) = 18 - 6 + 14 = 26$ (Lies on $P$).For $(4, 2, 2)$: $3(4) + 2 + 7(2) = 12 + 2 + 14 = 28 
eq 26$ (Does $NOT$ lie on $P$).For $(-8, 8, 6)$: $3(-8) + 8 + 7(6) = -24 + 8 + 42 = 26$ (Lies on $P$).

Let $P$ be the plane passing through the point $(1, 2, 3)$ and the line of intersection of the planes $\vec{r} \cdot (\hat{i} + \hat{j} + 4\hat{k}) = 16$ and $\vec{r} \cdot (-\hat{i} + \hat{j} + \hat{k}) = 6$. Then which of the following points does $NOT$ lie on $P$?

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