6. ⇒ (MHT CET 2023 10th May Evening Shift )

The plane through the intersection of planes $x + y + z = 1$ and $2 x + 3 y - z + 4 = 0$ and parallel to $Y$ -axis also passes through the point

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

B. $(- 3, 0, 1)$

C. $(3, 2, 1)$

D. $(- 3, 0, - 1)$

Correct answer option is (C)

Equation of plane through the intersection of given planes is

$\begin{aligned} (x + y + z - 1) + λ (2 x + 3 y - z + 4) = 0 \dots (i) \\ \Rightarrow (1 + 2 λ) x + (1 + 3 λ) y + (1 - λ) z + 4 λ - 1 = 0 \end{aligned}$

Since the plane is parallel to $Y$ -axis.

$\begin{array}{r} ∴ 1 + 3 λ = 0 \\ \Rightarrow λ = \frac{- 1}{3} \end{array}$

Substituting $λ = \frac{- 1}{3}$ in (i), we get

$\begin{aligned} (x + y + z - 1) - \frac{1}{3} (2 x + 3 y - z + 4) = 0 \\ \Rightarrow x + 4 z - 7 = 0 \end{aligned}$

Point $(3, 2, 1)$ satisfies this equation.

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

Let $P$ be a plane passing through the points $(2, 1, 0), (4, 1, 1)$ and $(5, 0, 1)$ and $R$ be the point $(2, 1, 6)$ . Then image of $R$ in the plane $P$ is

A. $(6, 5, 2)$

B. $(4, 3, 2)$

C. $(6, 5, - 2)$

D. $(3, 4, - 2)$

Correct answer option is (C)

Equation of the plane passing through $(2, 1, 0), (4, 1, 1)$ and $(5, 0, 1)$ is

$\begin{aligned} | \begin{array}{lll} x - 2 & y - 1 & z - 0 \\ 4 - 2 & 1 - 1 & 1 - 0 \\ 5 - 2 & 0 - 1 & 1 - 0 \end{array} | = 0 \\ \Rightarrow x + y - 2 z = 3 \end{aligned}$

$\begin{array}{ll} R^{'} (x, y, z) is image of R (2, 1, 6) w.r.t. to plane \\ x + y - 2 z = 3 \\ \Rightarrow & \frac{x - 2}{1} = \frac{y - 1}{1} = \frac{z - 6}{- 2} = \frac{- 2 [2 + 1 - 2 (6) - 3]}{1 + 1 + 4} \\ \Rightarrow & \frac{x - 2}{1} = \frac{y - 1}{1} = \frac{z - 6}{- 2} = 4 \\ \Rightarrow & x = 6, y = 5, z = - 2 \\ ∴ & R^{'} (x, y, z) \equiv (6, 5, - 2) \end{array}$

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

$ABC$ is a triangle in a plane with vertices $A (2, 3, 5), B (- 1, 3, 2)$ and $C (λ, 5, μ)$ . If median through $A$ is equally inclined to the co-ordinate axes, then value of $λ + μ$ is

A. 17

B. 10

C. 7

D. 3

Correct answer option is (A)

Let $A D$ be the median

$∴$ Co-ordinates of

$\begin{aligned} D \equiv (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}, \frac{z_{1} + z_{2}}{2}) \\ D \equiv (\frac{λ - 1}{2}, 4, \frac{μ + 2}{2}) \\ ∴ & \overset{―}{AD} = (\frac{λ - 1}{2} - 2) \hat{i} + (4 - 3) \hat{j} + (\frac{μ + 2}{2} - 5) \hat{k} \\ ∴ & \overset{―}{AD} = (\frac{λ - 5}{2}) \hat{i} + \hat{j} + (\frac{μ - 8}{2}) \hat{k} \end{aligned}$

Since $A D$ makes equal angle with co-ordinate axes, the direction ratios are equal.

$∴ \frac{λ - 5}{2} = 1 = \frac{μ - 8}{2}$

Consider,

$\begin{aligned} \frac{λ - 5}{2} = 1 \\ \Rightarrow λ - 5 = 2 \end{aligned}$

$\begin{aligned} \Rightarrow λ = 7 \\ and \frac{μ - 8}{2} = 1 \\ \Rightarrow μ - 8 = 2 \\ \Rightarrow μ = 10 \\ ∴ & λ + μ = 7 + 10 = 17 \end{aligned}$

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

The equation of a plane, containing the line of intersection of the planes $2 x - y - 4 = 0$ and $y + 2 z - 4 = 0$ and passing through the point $(2, 1, 0)$ , is

A. $3 x - 2 y + z = 4$

B. $3 x + 2 y + z = 4$

C. $3 x - 2 y - z = 4$

D. $3 x + 2 y - z = - 4$

Correct answer option is (C)

Equation of plane passing through the intersection of given planes is

$2 x - y - 4 + λ (y + 2 z - 4) = 0$ .... (i)

Since, the plane passes through $(2, 1, 0)$

$\begin{aligned} 2 (2) - 1 - 4 + λ (1 + 2 (0) - 4) = 0 \\ 4 - 1 - 4 - 3 λ = 0 \\ - 1 - 3 λ = 0 \\ λ = \frac{- 1}{3} \end{aligned}$

Substituting $λ = \frac{- 1}{3}$ in equation (i), we get

$\begin{aligned} 2 x - y - 4 - \frac{1}{3} (y + 2 z - 4) = 0 \\ 6 x - 3 y - 12 - y - 2 z + 4 = 0 \\ 6 x - 4 y - 2 z - 8 = 0 \\ 3 x - 2 y - z = 4 \end{aligned}$

10. ⇒ (MHT CET 2023 9th May Morning Shift )

The foot of the perpendicular drawn from the origin to the plane is $(4, - 2, 5)$ , then the Cartesian equation of the plane is

A. $4 x - 2 y + 5 z = 45$

B. $- 4 x + 2 y + 5 z = 45$

C. $4 x - 2 y + 5 z + 45 = 0$

D. $4 x + 2 y - 5 z + 45 = 0$

Correct answer option is (A)

Substitute $x = 4, y = - 2, z = 5$ in all options.

$∴$ For option $A$

$\begin{aligned} 4 x - 2 y + 5 z & = 4 (4) - 2 (- 2) + 5 (5) \\ = 16 + 4 + 25 \\ = 45 \end{aligned}$

$∴$ Only option A is satisfied by $(4, - 2, 5)$

⇒Line and Plane