1. ⇒ (MHT CET 2023 10th May Morning Shift )

The line $\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}$ lies in the plane $x + 3 y - α z + β = 0$ , then the value of $α^{2} + α β + β^{2}$ is

A. 127

B. 43

C. 109

D. 61

Correct answer option is (B)

Line $\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}$ lies in the plane $x + 3 y - α z + β = 0$

The direction ratios of the line are $3, - 5, 2$ .

The direction ratios of the normal to the plane are $1, 3, - α$ .

The given line is perpendicular to the normal of plane.

$\begin{aligned} ∴ & 3 (1) + (- 5) (3) + 2 (- α) = 0 \\ \Rightarrow 3 - 15 - 2 α = 0 \\ \Rightarrow - 12 - 2 α = 0 \\ \Rightarrow α = - 6 \end{aligned}$

Also, point $(2, 1, - 2)$ lies on the plane

$\begin{aligned} x + 3 y - α z + β = 0 \\ \Rightarrow 2 + 3 - (- 6) (- 2) + β = 0 \\ \Rightarrow 2 + 3 - 12 + β = 0 \\ \Rightarrow β = 7 \\ ∴ & α^{2} + α β + β^{2} = 36 - 42 + 49 = 43 \end{aligned}$

2. ⇒ (MHT CET 2023 9th May Evening Shift )

If a line $L$ is the line of intersection of the planes $2 x + 3 y + z = 1$ and $x + 3 y + 2 z = 2$ . If line $L$ makes an angle $α$ with the positive $X$ -axis, then the value of $\sec α$ is

A. $\sqrt{3}$

B. 2

C. 1

D. $\sqrt{2}$

Correct answer option is (A)

Given equations:

$\begin{aligned} 2 x + 3 y + z = 1 \\ 2 x + 3 y = 1 - z ...... (i) \\ x + 3 y + 2 z = 2 \\ x + 3 y = 2 - 2 z ...... (ii) \end{aligned}$

Subtracting (ii) from (i), we get

$\begin{aligned} 2 x + 3 y - x - 3 y = 1 - z - 2 + 2 z \\ x = - 1 + z \\ z = \frac{x + 1}{1} ...... (iii) \end{aligned}$

Putting value of $x$ in equation (ii), we get

$\begin{aligned} - 1 + z + 3 y = 2 - 2 z \\ 3 z = 3 - 3 y \\ z = \frac{y - 1}{- 1} ...... (iv) \end{aligned}$

From (iii), (iv)

$\frac{x + 1}{1} = \frac{y - 1}{- 1} = \frac{z}{1}$

Thus, angle between above line and $X$ -axis having Direction Ratio's $(1, 0, 0)$ is given as

$\begin{aligned} \cos α & = | \frac{1 \cdot (1) + 0 + 0}{\sqrt{1 + 1 + 1} \cdot \sqrt{1}} | = \frac{1}{\sqrt{3}} \\ ∴ \sec α & = \sqrt{3} \end{aligned}$

3. ⇒ (MHT CET 2023 9th May Evening Shift )

The co-ordinates of the point, where the line $\frac{x - 1}{2} = \frac{y - 2}{- 3} = \frac{z + 5}{4}$ meets the plane $2 x + 4 y - z = 3$ , are

A. $(3, - 1, - 1)$

B. $(3, 1, - 1)$

C. $(3, - 1, 1)$

D. $(- 3, - 1, - 1)$

Correct answer option is (A)

Given line is

$\begin{array}{ll} \frac{x - 1}{2} = \frac{y - 2}{- 3} = \frac{z + 5}{4} \\ Let \frac{x - 1}{2} = \frac{y - 2}{- 3} = \frac{z + 5}{4} = λ \\ ∴ & x - 1 = 2 λ, y - 2 = - 3 λ, z + 5 = 4 λ \\ x = 2 λ + 1, y = - 3 λ + 2, z = 4 λ - 5 \\ ∴ & 2 x + 4 y - z = 3 \\ \Rightarrow 2 (2 λ + 1) + 4 (- 3 λ + 2) - (4 λ - 5) = 3 \\ \Rightarrow 4 λ + 2 - 12 λ + 8 - 4 λ + 5 = 3 \\ \Rightarrow - 12 λ = 3 - 15 \\ ∴ & x = 3, y = - 1, z = - 1, \\ ∴ & Required co-ordinates are: (3, - 1, - 1) \end{array}$

4. ⇒ (MHT CET 2021 20th September Morning Shift )

If the line $\frac{x + 1}{2} = \frac{y - m}{3} = \frac{z - 4}{6}$ lies in the plane $3 x - 14 y + 6 z + 49 = 0$ , then the value of $m$ is

A. 3

B. -5

C. 5

D. 2

Correct answer option is (C)

Line $\frac{x + 1}{2} = \frac{y - m}{3} = \frac{z - 4}{6}$ lies in plane $3 x - 14 y + 6 z + 49 = 0$

$∴$ Point $(- 1, m, 4)$ lies in the plane.

$\begin{aligned} ∴ 3 (- 1) - 14 (m) + 6 (4) + 49 = 0 \Rightarrow - 3 - 14 m + 24 + 49 = 0 \\ ∴ 14 m = 70 \Rightarrow m = 5 \end{aligned}$

⇒Line and Plane