The equation of plane containing the line x-13=y+64=z+12 and parallel to the line x-22=y-1-3=z+45 is

The given lines are x-13=y+64=z+12 and x-22=y-1-3=z+45Since 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 linesHence, the normal vector to the plane is n1⇀×n2⇀ where n1⇀=3i+4j+2k and n2⇀=2i-3j+5kn1⇀×n2⇀=ijk3422-35 =i20+6-j15-4+k-9-8 =26i-11j-17kEquation of the plane passing through the point 1,-6,-1 and is perpendicular to the vector 26i-11j-17k is 26x-1-11y+6-17z+1=026x-11y-17z-26-66-17=026x-11y-17z-109=0

The equation of plane containing the line x-13=y+64=z+12 and parallel to the line x-22=y-1-3=z+45 is

The given lines are x-13=y+64=z+12 and x-22=y-1-3=z+45Since 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 linesHence, the normal vector to the plane is n1⇀×n2⇀ where n1⇀=3i+4j+2k and n2⇀=2i-3j+5kn1⇀×n2⇀=ijk3422-35 =i20+6-j15-4+k-9-8 =26i-11j-17kEquation of the plane passing through the point 1,-6,-1 and is perpendicular to the vector 26i-11j-17k is 26x-1-11y+6-17z+1=026x-11y-17z-26-66-17=026x-11y-17z-109=0

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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

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