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(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
\[
3\alpha-\gamma+2\beta=
\]
Show 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.
Updated On: Jul 9, 2026
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The Correct Option is C
Solution and Explanation
Concept:
If a point is reflected in a plane, then the plane bisects the segment joining the point and its image at right angles.
Since the plane passes through \(A\) and is perpendicular to \(BC\), the normal vector of the plane is parallel to \(BC\).
Step 1: Find the equation of the plane. \[ B(-1,6,2),\qquad C(1,1,-2). \] \[ \overrightarrow{BC} = (1+1,\;1-6,\;-2-2) = (2,-5,-4). \] Hence a normal vector to the plane is \[ \vec n=(2,-5,-4). \] Since the plane passes through \[ A(3,1,2), \] its equation is \[ 2(x-3)-5(y-1)-4(z-2)=0. \] \[ 2x-5y-4z+7=0. \]
Step 2: Find the image of \(C(1,1,-2)\) in the plane. For the plane \[ ax+by+cz+d=0, \] the reflection of \((x_0,y_0,z_0)\) is \[ \left( x_0-\frac{2aD}{a^2+b^2+c^2}, y_0-\frac{2bD}{a^2+b^2+c^2}, z_0-\frac{2cD}{a^2+b^2+c^2} \right), \] where \[ D=ax_0+by_0+cz_0+d. \] Here \[ a=2,\quad b=-5,\quad c=-4,\quad d=7. \] For \(C(1,1,-2)\), \[ D=2(1)-5(1)-4(-2)+7. \] \[ D=2-5+8+7=12. \] Also, \[ a^2+b^2+c^2 = 4+25+16 = 45. \] Hence \[ \alpha = 1-\frac{2(2)(12)}{45} = 1-\frac{16}{15} = -\frac1{15}, \] \[ \beta = 1-\frac{2(-5)(12)}{45} = 1+\frac{8}{3} = \frac{11}{3}, \] \[ \gamma = -2-\frac{2(-4)(12)}{45} = -2+\frac{32}{15} = \frac{2}{15}. \]
Step 3: Evaluate \(3\alpha-\gamma+2\beta\). \[ 3\alpha-\gamma+2\beta = 3\left(-\frac1{15}\right) -\frac2{15} +2\left(\frac{11}{3}\right). \] \[ = -\frac3{15} -\frac2{15} +\frac{22}{3}. \] \[ = -\frac5{15} +\frac{22}{3}. \] \[ = -\frac13+\frac{22}{3}. \] \[ = \frac{21}{3}. \] \[ =7. \]
Step 4: Write the final answer. \[ \boxed{7} \]
Step 1: Find the equation of the plane. \[ B(-1,6,2),\qquad C(1,1,-2). \] \[ \overrightarrow{BC} = (1+1,\;1-6,\;-2-2) = (2,-5,-4). \] Hence a normal vector to the plane is \[ \vec n=(2,-5,-4). \] Since the plane passes through \[ A(3,1,2), \] its equation is \[ 2(x-3)-5(y-1)-4(z-2)=0. \] \[ 2x-5y-4z+7=0. \]
Step 2: Find the image of \(C(1,1,-2)\) in the plane. For the plane \[ ax+by+cz+d=0, \] the reflection of \((x_0,y_0,z_0)\) is \[ \left( x_0-\frac{2aD}{a^2+b^2+c^2}, y_0-\frac{2bD}{a^2+b^2+c^2}, z_0-\frac{2cD}{a^2+b^2+c^2} \right), \] where \[ D=ax_0+by_0+cz_0+d. \] Here \[ a=2,\quad b=-5,\quad c=-4,\quad d=7. \] For \(C(1,1,-2)\), \[ D=2(1)-5(1)-4(-2)+7. \] \[ D=2-5+8+7=12. \] Also, \[ a^2+b^2+c^2 = 4+25+16 = 45. \] Hence \[ \alpha = 1-\frac{2(2)(12)}{45} = 1-\frac{16}{15} = -\frac1{15}, \] \[ \beta = 1-\frac{2(-5)(12)}{45} = 1+\frac{8}{3} = \frac{11}{3}, \] \[ \gamma = -2-\frac{2(-4)(12)}{45} = -2+\frac{32}{15} = \frac{2}{15}. \]
Step 3: Evaluate \(3\alpha-\gamma+2\beta\). \[ 3\alpha-\gamma+2\beta = 3\left(-\frac1{15}\right) -\frac2{15} +2\left(\frac{11}{3}\right). \] \[ = -\frac3{15} -\frac2{15} +\frac{22}{3}. \] \[ = -\frac5{15} +\frac{22}{3}. \] \[ = -\frac13+\frac{22}{3}. \] \[ = \frac{21}{3}. \] \[ =7. \]
Step 4: Write the final answer. \[ \boxed{7} \]
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