The question asks to derive, prove, or verify the Cartesian equation (ax + by + cz = d) of a specified plane, typically defined by three points or geometric conditions.
In coordinate space, the equation of the plane that is perpendicular to the line $\frac { x - 2 } { 2 } = \frac { y - 2 } { 3 } = z - 1$ and passes through the point $( 1 , - 5,2 )$ is $2 x + a y + b z + c = 0$. Find the value of $a + b + c$. [3 points]
The point $L$, which lies vertically above the midpoint of the edge $\left[ \mathrm { A } _ { 1 } \mathrm {~A} _ { 2 } \right]$, represents the position of a floodlight in the model, which is installed 12 m above the base. The points $L , B _ { 2 }$ and $B _ { 3 }$ determine a plane $F$. Find an equation of $F$ in normal form. (for verification: $F : 3 x _ { 1 } + x _ { 2 } + 5 x _ { 3 } - 90 = 0$ )
135- The plane passing through the two intersecting lines $(D): \begin{cases} 2x+y=3 \\ 2y-z=0 \end{cases}$ and $(D'): \dfrac{x+1}{2}=\dfrac{y}{1}=\dfrac{z+1}{3}$. Which value does the $z$-axis intercept cut? (1) $-0.8$ (2) $-0.6$ (3) $0.8$ (4) $1.2$
135- The plane passing through the line with equation $\dfrac{x+1}{2} = \dfrac{y}{3} = \dfrac{z-2}{-1}$ and the point $(0,3,0)$ intersects the $Z$-axis at what elevation? (1) $-2$ (2) $-3$ (3) $2$ (4) $3$
Equation of the plane containing the straight line $\frac { x } { 2 } = \frac { y } { 3 } = \frac { z } { 4 }$ and perpendicular to the plane containing the straight lines $\frac { x } { 3 } = \frac { y } { 4 } = \frac { z } { 2 }$ and $\frac { x } { 4 } = \frac { y } { 2 } = \frac { z } { 3 }$ is A) $x + 2 y - 2 z = 0$ B) $3 x + 2 y - 2 z = 0$ C) $x - 2 y + z = 0$ D) $5 x + 2 y - 4 z = 0$
Consider the lines $L _ { 1 } : \frac { x - 1 } { 2 } = \frac { y } { - 1 } = \frac { z + 3 } { 1 } , L _ { 2 } : \frac { x - 4 } { 1 } = \frac { y + 3 } { 1 } = \frac { z + 3 } { 2 }$ and the planes $P _ { 1 } : 7 x + y + 2 z = 3 , P _ { 2 } : 3 x + 5 y - 6 z = 4$. Let $a x + b y + c z = d$ be the equation of the plane passing through the point of intersection of lines $L _ { 1 }$ and $L _ { 2 }$, and perpendicular to planes $P _ { 1 }$ and $P _ { 2 }$. Match List I with List II and select the correct answer using the code given below the lists: List I
Equation of the plane which passes through the point of intersection of lines $\frac { x - 1 } { 3 } = \frac { y - 2 } { 1 } = \frac { z - 3 } { 2 }$ and $\frac { x - 3 } { 1 } = \frac { y - 1 } { 2 } = \frac { z - 2 } { 3 }$ and has the largest distance from the origin is: (1) $4 x + 3 y + 5 z = 50$ (2) $3 x + 4 y + 5 z = 49$ (3) $5 x + 4 y + 3 z = 57$ (4) $7 x + 2 y + 4 z = 54$
The equation of the plane containing the line of intersection of $2 x - 5 y + z = 3 ; x + y + 4 z = 5$, and parallel to the plane, $x + 3 y + 6 z = 1$, is (1) $2 x + 6 y + 12 z = - 13$ (2) $2 x + 6 y + 12 z = 13$ (3) $x + 3 y + 6 z = - 7$ (4) $x + 3 y + 6 z = 7$
The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$ is (1) $\vec { r } \times ( \hat { i } + \hat { k } ) + 2 = 0$ (2) $\vec { r } \cdot ( \hat { i } - \hat { k } ) - 2 = 0$ (3) $\vec { r } \times ( \hat { i } - \hat { k } ) + 2 = 0$ (4) $\vec { r } \cdot ( \hat { i } - \hat { k } ) + 2 = 0$
The equation of a plane containing the line of intersection of the planes $2x - y - 4 = 0$ and $y + 2z - 4 = 0$ and passing through the point $(1,1,0)$ is (1) $x - 3y - 2z = -2$ (2) $x + 3y + z = 4$ (3) $x - y - z = 0$ (4) $2x - z = 2$
The plane passing through the points $(1, 2, 1)$, $(2, 1, 2)$ and parallel to the line, $2x = 3y, z = 1$ also passes through the point (1) $(0, 6, -2)$ (2) $(-2, 0, 1)$ (3) $(0, -6, 2)$ (4) $(2, 0, -1)$
If the equation of plane passing through the mirror image of a point $( 2,3,1 )$ with respect to line $\frac { x + 1 } { 2 } = \frac { y - 3 } { 1 } = \frac { z + 2 } { - 1 }$ and containing the line $\frac { x - 2 } { 3 } = \frac { 1 - y } { 2 } = \frac { z + 1 } { 1 }$ is $\alpha x + \beta y + \gamma z = 24$ then $\alpha + \beta + \gamma$ is equal to: (1) 20 (2) 19 (3) 18 (4) 21
Let the equation of the plane, that passes through the point $( 1,4 , - 3 )$ and contains the line of intersection of the planes $3 x - 2 y + 4 z - 7 = 0$ and $x + 5 y - 2 z + 9 = 0$, be $\alpha x + \beta y + \gamma z + 3 = 0$, then $\alpha + \beta + \gamma$ is equal to : (1) $- 15$ (2) 15 (3) $- 23$ (4) 23
If the plane $P$ passes through the intersection of two mutually perpendicular planes $2 x + k y - 5 z = 1$ and $3 k x - k y + z = 5 , k < 3$ and intercepts a unit length on positive $x$-axis, then the intercept made by the plane $P$ on the $y$-axis is (1) $\frac { 1 } { 11 }$ (2) $\frac { 5 } { 11 }$ (3) 6 (4) 7
Let the equation of the plane passing through the line $x - 2 y - z - 5 = 0 = x + y + 3 z - 5$ and parallel to the line $x + y + 2 z - 7 = 0 = 2 x + 3 y + z - 2$ be $a x + b y + c z = 65$. Then the distance of the point $( a , b , c )$ from the plane $2 x + 2 y - z + 16 = 0$ is $\_\_\_\_$.
Let the equation of the plane P containing the line $x + 10 = \frac { 8 - y } { 2 } = z$ be $a x + b y + 3 z = 2 ( a + b )$ and the distance of the plane P from the point $( 1,27,7 )$ be $c$. Then $\mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } + \mathrm { c } ^ { 2 }$ is equal to
Let the plane $P : 4 x - y + z = 10$ be rotated by an angle $\frac { \pi } { 2 }$ about its line of intersection with the plane $x + y - z = 4$. If $\alpha$ is the distance of the point $( 2,3 , - 4 )$ from the new position of the plane $P$, then $35 \alpha$ is equal to (1) 85 (2) 105 (3) 126 (4) 90
In coordinate space, there are two lines $L _ { 1 } , L _ { 2 }$ and a plane $E$. The line $L _ { 1 } : \frac { x } { 2 } = \frac { y } { - 3 } = \frac { z } { - 5 }$, and the parametric equation of $L _ { 2 }$ is $\left\{ \begin{array} { l } x = 1 \\ y = 1 + 2 t \\ z = 1 + 3 t \end{array} \right.$ ($t$ is a real number). If $L _ { 1 }$ lies on $E$ and $L _ { 2 }$ does not intersect $E$, then the equation of $E$ is $x -$ (16) $y +$ (17) $z =$ (18).