QJEEM23Mac6e5: JEE Main 2023 Mathematics - Class 12 JEEM23Mac6e5 Solution: The line, that is coplanar to the line , is :

Given two lines:

$\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}$

and

$\frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}$

These lines are coplanar if the determinant of the matrix

$\begin{vmatrix}x_2-x_1 & y_2-y_1 & z_2-z_1 \\a_1 & b_1 & c_1 \\a_2 & b_2 & c_2\end{vmatrix}$ = 0

Now let's apply this condition to the given problem. The given line is :

$\frac{x + 3}{-3} = \frac{y - 1}{1} = \frac{z - 5}{5}$

So, the coordinates of any point on this line are $(-3, 1, 5)$ and the direction ratios are $(-3, 1, 5)$ .

Now, let's calculate the determinants for each option and check which one equals zero.

For Option A :

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

The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(-1, 2, 4)$ .

The determinant is :

$\begin{vmatrix}-1 - (-3) & 2 - 1 & 5 - 5 \\-3 & 1 & 5 \\-1 & 2 & 4\end{vmatrix}=\begin{vmatrix}2 & 1 & 0 \\-3 & 1 & 5 \\-1 & 2 & 4\end{vmatrix}$

Applying the formula :

= $2(1 \times 4 - 5\times2) - 1((-3\times4) - (5\times-1)) + 0((-3\times2) - (1\times-1))$

= $2(4 - 10) - 1(-12 - (-5)) + 0(-6 - (-1))$

= $2(-6) - 1(-7) + 0(-5)$

= $-12 + 7 + 0$

= $-5$

The determinant for Option A is not equal to zero, so this line is not coplanar with the given line.

For Option B, we have :

$\frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}$

The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(-1, 2, 5)$ .

$\begin{vmatrix}-1 - (-3) & 2 - 1 & 5 - 5 \\-3 & 1 & 5 \\-1 & 2 & 5\end{vmatrix}=\begin{vmatrix}2 & 1 & 0 \\-3 & 1 & 5 \\-1 & 2 & 5\end{vmatrix}$

Applying the formula, we get :

= $2(1*5 - 5*2) - 1((-3*5) - (5*-1)) + 0((-3*2) - (1*-1))$

= $2(5 - 10) - 1(-15 - (-5)) + 0(-6 - (-1))$

= $2(-5) - 1(-10) + 0(-5)$

= $-10 + 10 + 0$

= $0$

So, the determinant for Option B equals zero, which confirms that the line in Option B is coplanar with the given line.

For Option C :

$\frac{x - 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}$

The coordinates of any point on this line are $(1, 2, 5)$ and the direction ratios are $(-1, 2, 5)$ .

The determinant is:

$\begin{vmatrix}1 - (-3) & 2 - 1 & 5 - 5 \\-3 & 1 & 5 \\-1 & 2 & 5\end{vmatrix}=\begin{vmatrix}4 & 1 & 0 \\-3 & 1 & 5 \\-1 & 2 & 5\end{vmatrix}$

Applying the formula :

= $4(1*5 - 5*2) - 1((-3*5) - (5*-1)) + 0((-3*2) - (1*-1))$

= $4(5 - 10) - 1(-15 - (-5)) + 0(-6 - (-1))$

= $4(-5) - 1(-10) + 0(-5)$

= $-20 + 10 + 0$

= $-10$

The determinant for Option C is not equal to zero, so this line is not coplanar with the given line.

For Option D :

$\frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z - 5}{5}$

The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(1, 2, 5)$ .

The determinant is :

$\begin{vmatrix}-1 - (-3) & 2 - 1 & 5 - 5 \\-3 & 1 & 5 \\1 & 2 & 5\end{vmatrix}=\begin{vmatrix}2 & 1 & 0 \\-3 & 1 & 5 \\1 & 2 & 5\end{vmatrix}$

Applying the formula :

= $2(1*5 - 5*2) - 1((-3*5) - (5*1)) + 0((-3*2) - (1*1))$

= $2(5 - 10) - 1(-15 - 5) + 0(-6 - 1)$

= $2(-5) - 1(-20) + 0(-7)$

= $-10 + 20 + 0$

= $10$

So the determinant for Option D is not equal to zero, which means the line in Option D is not coplanar with the given line.