train/precalculus/1050.json

{
    "problem": "Line $L$ is the intersection of the planes $x + 2y + 3z = 2$ and $x - y + z = 3.$  A plane $P,$ different from both these planes, contains line $L,$ and has a distance of $\\frac{2}{\\sqrt{3}}$ from the point $(3,1,-1).$  Find the equation of plane $P.$  Enter your answer in the form\n\\[Ax + By + Cz + D = 0,\\]where $A,$ $B,$ $C,$ $D$ are integers such that $A > 0$ and $\\gcd(|A|,|B|,|C|,|D|) = 1.$",
    "level": "Level 5",
    "type": "Precalculus",
    "solution": "We can write the equations of the planes as $x + 2y + 3z - 2 = 0$ and $x - y + z - 3 = 0.$  Any point in $L$ satisfies both equations, which means any point in $L$ satisfies an equation of the form\n\\[a(x + 2y + 3z - 2) + b(x - y + z - 3) = 0.\\]We can write this as\n\\[(a + b)x + (2a - b)y + (3a + b)z - (2a + 3b) = 0.\\]The distance from this plane to $(3,1,-1)$ is $\\frac{2}{\\sqrt{3}}.$  Using the formula for the distance from a point to a plane, we get\n\\[\\frac{|(a + b)(3) + (2a - b)(1) + (3a + b)(-1) - (2a + 3b)|}{\\sqrt{(a + b)^2 + (2a - b)^2 + (3a + b)^2}} = \\frac{2}{\\sqrt{3}}.\\]We can simplify this to\n\\[\\frac{|2b|}{\\sqrt{14a^2 + 4ab + 3b^2}} = \\frac{2}{\\sqrt{3}}.\\]Then $|b| \\sqrt{3} = \\sqrt{14a^2 + 4ab + 3b^2}.$  Squaring both sides, we get $3b^2 = 14a^2 + 4ab + 3b^2,$ so\n\\[14a^2 + 4ab = 0.\\]This factors as $2a(7a + 2b) = 0.$  If $a = 0,$ then plane $P$ will coincide with the second plane $x - y + z = 3.$  So, $7a + 2b = 0.$  We can take $a = 2$ and $b = -7,$ which gives us\n\\[(2)(x + 2y + 3z - 2) + (-7)(x - y + z - 3) = 0.\\]This simplifies to $\\boxed{5x - 11y + z - 17 = 0}.$"
}