1. ⇒ (MHT CET 2023 12th May Evening Shift )

The equation of the line, passing through $(1, 2, 3)$ and parallel to planes $x - y + 2 z = 5$ and $3 x + y + z = 6$ , is

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

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

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

D. $\frac{x - 1}{5} = \frac{y - 2}{7} = \frac{z - 3}{1}$

Correct answer option is (A)

Required equation of line is

$\begin{aligned} \frac{x - 1}{| \begin{array}{cc} - 1 & 2 \\ 1 & 1 \end{array} |} = \frac{y - 2}{- | \begin{array}{cc} 1 & 2 \\ 3 & 1 \end{array} |} = \frac{z - 3}{| \begin{array}{cc} 1 & - 1 \\ 3 & 1 \end{array} |} \\ ∴ \frac{x - 1}{- 3} = \frac{y - 2}{5} = \frac{z - 3}{4} \end{aligned}$

2. ⇒ (MHT CET 2023 11th May Evening Shift )

$x - 3 = \frac{y - k}{2} = z$ intersect, then the value of $k$ is

A. $\frac{3}{2}$

B. $\frac{- 2}{9}$

C. $\frac{- 2}{3}$

D. $\frac{9}{2}$

Correct answer option is (D)

As the given lines are intersecting, the shortest distance between them is zero.

$\begin{aligned} ∴ | \begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array} | = 0 \\ ∴ | \begin{array}{ccc} 3 - 1 & k + 1 & 0 - 1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array} | = 0 \\ ∴ | \begin{array}{ccc} 2 & k + 1 & - 1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array} | = 0 \\ ∴ k = \frac{9}{2} \end{aligned}$

3. ⇒ (MHT CET 2023 11th May Evening Shift )

The equation of line passing through the point $(1, 2, 3)$ and perpendicular to the lines $\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z + 1}{- 2}$ and $\frac{x}{2} = \frac{y}{- 3} = \frac{z}{1}$ is

A. $\overset{―}{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (4 \hat{i} + 7 \hat{j} - 13 \hat{k})$

B. $\overset{―}{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (- 4 \hat{i} + 7 \hat{j} - 13 \hat{k})$

C. $\overset{―}{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (- 4 \hat{i} - 7 \hat{j} - 13 \hat{k})$

D. $\overset{―}{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (4 \hat{i} - 7 \hat{j} - 13 \hat{k})$

Correct answer option is (C)

Required line is perpendicular to the lines $\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z + 1}{- 2}$ and $\frac{x}{2} = \frac{y}{- 3} = \frac{z}{1}$

$∴$ Required line is parallel to vector

$\bar{b} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & - 2 \\ 2 & - 3 & 1 \end{array} | = - 4 \hat{i} - 7 \hat{j} - 13 \hat{k}$

$∴$

The equation of the required line is $(\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (- 4 \hat{i} - 7 \hat{j} - 13 \hat{k})$

4. ⇒ (MHT CET 2023 11th May Morning Shift )

If the direction cosines $l, m, n$ of two lines are connected by relations $l - 5 m + 3 n = 0$ and $7 l^{2} + 5 m^{2} - 3 n^{2} = 0$ , then value of $l + m + n$ is

A. $\frac{2}{\sqrt{6}}$ or $\frac{6}{\sqrt{14}}$

B. $\frac{1}{\sqrt{6}}$ or $\frac{5}{\sqrt{14}}$

C. $\frac{2}{\sqrt{6}}$ or $\frac{5}{\sqrt{14}}$

D. $\frac{1}{\sqrt{6}}$ or $\frac{6}{\sqrt{14}}$

Correct answer option is (A)

$\begin{aligned} l - 5 m + 3 n = 0 and 7 l^{2} + 5 m^{2} - 3 n^{2} = 0 \\ \Rightarrow l = 5 m - 3 n and 7 l^{2} = 3 n^{2} - 5 m^{2} \end{aligned}$

$\begin{aligned} \Rightarrow l = 5 m - 3 n and 7 l^{2} = 3 n^{2} - 5 m^{2} \\ Putting l = (5 m - 3 n) in 7 l^{2} = 3 n^{2} - 5 m^{2}, we \\ 7 (5 m - 3 n)^{2} = 3 n^{2} - 5 m^{2} \\ \Rightarrow 7 (25 m^{2} - 30 m n + 9 n^{2}) = 3 n^{2} - 5 m^{2} \\ \Rightarrow 180 m^{2} - 210 m n + 60 n^{2} = 0 \\ \Rightarrow 6 m^{2} - 7 m n + 2 n^{2} = 0 \\ \Rightarrow (3 m - 2 n) (2 m - n) = 0 \end{aligned}$

$\Rightarrow 3 m = 2 n or 2 m = n$ ..... (i)

If $3 m = 2 n$ , then $l = \frac{n}{3}$

$\begin{aligned} ∴ \frac{m}{2} = \frac{n}{3} = \frac{l}{1} = \frac{1}{\sqrt{14}} \\ ∴ l + m + n = \frac{6}{\sqrt{14}} \end{aligned}$

If $2 m = n$ , then $l = \frac{- n}{2}$

$\begin{array}{ll} ∴ & \frac{m}{1} = \frac{n}{2} = \frac{l}{- 1} = \frac{1}{\sqrt{6}} \\ ∴ & l + m + n = \frac{2}{\sqrt{6}} \end{array}$

$∴$ The possible values of $l + m + n$ is $\frac{2}{\sqrt{6}}$ or $\frac{6}{\sqrt{14}}$

5. ⇒ (MHT CET 2023 11th May Morning Shift )

The vector equation of the line $2 x + 4 = 3 y + 1 = 6 z - 3$ is

A. $\overset{―}{r} = (2 \hat{i} + \frac{1}{3} \hat{j} + \frac{1}{2} \hat{k}) + λ (3 \hat{i} + 2 \hat{j} + \overset{―}{k})$

B. $\overset{―}{r} = (- 2 \hat{i} - \frac{1}{3} \hat{j} + \frac{1}{2} \hat{k}) + λ (3 \hat{i} + 2 \hat{j} + \hat{k})$

C. $\overset{―}{r} = (2 \hat{i} + \hat{j} + \hat{k}) + λ (3 \hat{i} + 2 \hat{j} + \overset{―}{k})$

D. $\overset{―}{r} = (- 2 \hat{i} - \hat{j} + \hat{k}) + λ (3 \hat{i} + 2 \hat{j} + \hat{k})$

Correct answer option is (B)

The equation of line is

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

$∴$ The given line passes through $(- 2, - \frac{1}{3}, \frac{1}{2})$ and has direction ratios proportional to $3, 2, 1$ .

$∴$ Vector equation of the line is

$\bar{r} = (- 2 \hat{i} - \frac{1}{3} \hat{j} + \frac{1}{2} \hat{k}) + λ (3 \hat{i} + 2 \hat{j} + \hat{k})$

⇒Line and Plane