For the first find an equation of intersection curve.
{4x2+y2+4z2+4yz+2zx=0x+2y+3z=0
{z=−3x+2y4x2+y2+4(3x+2y)2−34y(x+2y)−32x(x+2y)=0
34x2+y2−8xy=0
Transform the equation
(y−(4−32i)x)(y−(4+32i)x)=0
This is equation of two imaginary intersecting lines.
Angle between the lines y=k1x and y=k2x is tanα=1+k2k1k2−k1
In our case k1=4−32i and k2=4+32i .
It mean, that tanα=1+(4−32i)(4+32i)(4+32i)−(4−32i)=3462i=1732i
The angle between the lines of intersection is
α=tan−1(1732i)
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