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Question

Find the point where the line of intersection of the planes $$ x - 2y + z = 1$$ and $$x + 2y - 2z = 5$$, intersects the plane $$2x + 2y + z + 6 = 0$$


A
(1,2,4)
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B
(0,0,6)
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C
(1,0,8)
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D
(1,1,2)
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Solution

The correct option is B $$(1, -2, -4)$$
Given planes  are $$x - 2y + z = 1 ...(i)$$
$$x + 2y - 2z = 5 ...(ii)$$
and $$2x + 2y + z = -6 ...(iii)$$
Add $$(i) + (ii) +(iii)$$
$$4x + 2y = 0 \Rightarrow  y = -2x ...(iv)$$
From equations $$(iii) - (i)$$
$$x + 4y = -7 ...(v)$$
from (iv) and (v) we get
$$x = 1, y = -2$$
Put in(i) we get $$z = -4$$
So point of intersection is $$(1, -2, -4) $$

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