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)
B
(0,0,6)
C
(1,0,8)
D
(1,1,2)

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 $$(1, -2, -4)$$Maths

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