Equation of a Plane : Point Normal Form

Q. If direction ratios of the normal of the plane which contains the line $\frac{x-2}{3}=\frac{y-4}{2}=\frac{z-1}{1}$ is $(a,1,-26),$ then $a=$

Q. The equation of a plane passing through (1, 2, –3), (0, 0, 0) and perpendicular to the plane 3x – 5y + 2z = 11, is :

1. 4x + y + 2z = 0
2. x + y + z = 0
3. $3x+y+\frac{5}{3}z=0$
4. $3x-y+\frac{z}{3}=0$

Q. The equation of the plane passing through the point (–2, –2, 2) and containing the line joining the points (1, 1, 1) and (1, –1, 2) is :

1. x + 2y – 3z + 4 = 0
2. 5x + 2y – 3z – 17 = 0
3. x – 3y – 6z + 8 = 0
4. 3z – 4y + 1 = 0

Q. P is a fixed point (a, a, a) on a line through the origin equally inclined to the axes, then any plane through P perpendicular to OP, makes intercepts on the axes, the sum of whose reciprocals is equal to

1. a
2. $\frac{3}{2a}$
3. $\frac{3a}{2}$
4. $\frac{1}{a}$

Q. Equation of the plane perpendicular to the plane x – 2y + 5z + 1 = 0 which passes through the points (2, –3, 1) and (–1, 1, –7) is given by :

1. 4x – 4y + z + 7 = 0
2. 4x + 7y + 2z + 11 = 0
3. 2x + y – z = 0
4. 2x + y – 3z = 0

Q. The equation to the plane through the points (2, -1, 0), (3, -4, 5) parallel to a line with direction cosines proportional to (2, 3, 4 ) is 9x – 2y – 3z = k, where k is

1. 10
2. -10
3. 20
4. -20

Q. The equation of the plane passing through the points (2, –4, 0), (3, –4, 5) and parallel to the line 2x = 3y = 4z is :

1. 29x – 27y – 22z = 85
2. 73x + 61y – 22z = 85
3. 125x – 90y – 79z = 340
4. 32x – 21y – 36z = 85

Q.

The distance of the point (1, 3, -7) from the plane passing through the point (1, -1, -1) having normal perpendicular to both the lines
$\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$ and $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}$, is

1. $\frac{20}{\sqrt{74}}$ units

2. $\frac{10}{\sqrt{83}}$ units

3. $\frac{5}{\sqrt{83}}$ units

4. $\frac{10}{\sqrt{74}}$ units

Q.

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 $P_1:7x+y+2z=3,P_2:3x+5y-6z=4.$ Let $ax+by+cz=d$ 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.

$$\begin{array}{cccc}ListI& & ListII& \\ P& a=& 1& 13\\ Q& b=& 2& -3\\ R& c=& 3& 1\\ S& d=& 4& -2\end{array}$$

1. P-3, Q-2, R-4, S-1

2. P-2, Q-4, R-1, S-3

3. P-1, Q-3, R-4, S-2

4. P-3, Q-2, R-1, S-4

Q. If the point $(2,\alpha ,\beta )$ lies on the plane which passes through the point $(3,4,2)$ and$(7,0,6)$ and is perpendicular to the plane $2x-5y=15$, then $2\alpha -3\beta$ is equal to :

1. $17$
2. $12$
3. $5$
4. $7$