The equation of plane containing the line $\frac{x - 1}{3} = \frac{y + 6}{4} = \frac{z + 1}{2}$ and parallel to the line $\frac{x - 2}{2} = \frac{y - 1}{- 3} = \frac{z + 4}{5}$ is

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$\begin{array}{rcl} 26 x - 11 y + 17 z + 109 & = & 0 \end{array}$

$\begin{array}{rcl} 26 x + 11 y - 17 z - 109 & = & 0 \end{array}$

$\begin{array}{rcl} 26 x - 11 y - 17 z - 109 & = & 0 \end{array}$

$\begin{array}{rcl} 26 x - 11 y - 17 z + 109 & = & 0 \end{array}$

answer is C.

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Detailed Solution

The given lines are $\frac{x - 1}{3} = \frac{y + 6}{4} = \frac{z + 1}{2}$ and $\frac{x - 2}{2} = \frac{y - 1}{- 3} = \frac{z + 4}{5}$

Since first line lies on the required plane and the second line is parallel to the required plane.

hence the normal vector is perpendicular to both vector along the given lines

Hence, the normal vector to the plane is $\overset{⇀}{n_{1}} \times \overset{⇀}{n_{2}}$ where $\overset{⇀}{n_{1}} = 3 i + 4 j + 2 k$ and $\overset{⇀}{n_{2}} = 2 i - 3 j + 5 k$

$\begin{array}{rcl} \overset{⇀}{n_{1}} \times \overset{⇀}{n_{2}} & = & |\begin{matrix} i & j & k \\ 3 & 4 & 2 \\ 2 & - 3 & 5 \end{matrix}| \\ = & i (20 + 6) - j (15 - 4) + k (- 9 - 8) \\ = & 26 i - 11 j - 17 k \end{array}$

Equation of the plane passing through the point $(1, - 6, - 1)$ and is perpendicular to the vector $26 i - 11 j - 17 k$ is

$\begin{array}{rcl} 26 (x - 1) - 11 (y + 6) - 17 (z + 1) & = & 0 \\ 26 x - 11 y - 17 z - 26 - 66 - 17 & = & 0 \\ 26 x - 11 y - 17 z - 109 & = & 0 \end{array}$