Question

A(3,1,2), B(-1,6,2) and C(1,1,-2) are three points. A plane passes through \(A\) and is perpendicular to the line joining \(B\) and \(C\). If \((\alpha,\beta,\gamma)\) is the image of \(C\) with respect to the plane, then \[ 3\alpha-\gamma+2\beta= \]

• A. \(0\)
• B. \(6\)
• C. \(7\)
• D. \(5\) \bigskip

Hint:

To reflect a point across a plane: \[ P' = P - \frac{2D}{a^2+b^2+c^2}(a,b,c), \] where \[ D=ax_0+by_0+cz_0+d. \] This formula directly gives the image coordinates.

Answer 1

The Correct Option is C

Concept: The plane is perpendicular to BC, so its normal is \(\overrightarrow{BC}\). Find the plane equation through A, then reflect C across the plane using the formula, and compute the required linear combination.

Step 1: B(-1,6,2), C(1,1,-2). \(\overrightarrow{BC} = (2,-5,-4)\). Plane through A(3,1,2): \(2(x-3)-5(y-1)-4(z-2)=0 \Rightarrow 2x-5y-4z+7=0\).

Step 2: Reflect C(1,1,-2). \(D = 2(1)-5(1)-4(-2)+7 = 2-5+8+7=12\). Denominator \(a^2+b^2+c^2=4+25+16=45\). Image: \(\alpha = 1 - \frac{2\cdot2\cdot12}{45} = 1 - \frac{48}{45} = -\frac{3}{45} = -\frac{1}{15}\). \(\beta = 1 - \frac{2\cdot(-5)\cdot12}{45} = 1 + \frac{120}{45} = 1 + \frac{8}{3} = \frac{11}{3}\). \(\gamma = -2 - \frac{2\cdot(-4)\cdot12}{45} = -2 + \frac{96}{45} = -2 + \frac{32}{15} = \frac{2}{15}\).

Step 3: \(3\alpha - \gamma + 2\beta = 3(-1/15) - 2/15 + 2(11/3) = -3/15 - 2/15 + 22/3 = -5/15 + 22/3 = -1/3 + 22/3 = 21/3 = 7\).

Step 4: Write the final answer. \(\boxed{7}\)