Assume the plane 4x−3y+7z=0 being horizontal, then the equation of the line of greatest slope (the line of greatest slope is the line which makes π2 with the line of intersection of the planes) through the point (2,−1,1) and lies in the plane 2x+y−3z=0 is

Let n¯1=4i−3j+7kn¯2=2i+j−3kn¯1×n¯2=2i+26j+10kDr’s of the line of intersection are (2,26,10) Required line is perpendicular to both n¯1×n¯2&n¯2Dr’s of the required line are -44,13,-25Equation of the required linex−2−44=y+113=z−1−25

Assume the plane 4x−3y+7z=0 being horizontal, then the equation of the line of greatest slope (the line of greatest slope is the line which makes π2 with the line of intersection of the planes) through the point (2,−1,1) and lies in the plane 2x+y−3z=0 is

Let n¯1=4i−3j+7kn¯2=2i+j−3kn¯1×n¯2=2i+26j+10kDr’s of the line of intersection are (2,26,10) Required line is perpendicular to both n¯1×n¯2&n¯2Dr’s of the required line are -44,13,-25Equation of the required linex−2−44=y+113=z−1−25

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Assume the plane $4 x - 3 y + 7 z = 0$ being horizontal, then the equation of the line of greatest slope (the line of greatest slope is the line which makes $\frac{π}{2}$ with the line of intersection of the planes) through the point $(2, - 1, 1)$ and lies in the plane $2 x + y - 3 z = 0$ is

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$\frac{x - 2}{13} = \frac{y + 1}{- 44} = \frac{z - 1}{- 25}$

$\frac{x - 2}{13} = \frac{y + 1}{- 25} = \frac{z - 1}{- 44}$

$\frac{x - 2}{- 44} = \frac{y + 1}{13} = \frac{z - 1}{- 25}$

$\frac{x - 2}{- 25} = \frac{y + 1}{13} = \frac{z - 1}{- 44}$

answer is C.

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Detailed Solution

Let ${\bar{n}}_{1} = 4 i - 3 j + 7 k$
${\bar{n}}_{2} = 2 i + j - 3 k$
${\bar{n}}_{1} \times {\bar{n}}_{2} = 2 i + 26 j + 10 k$
Dr’s of the line of intersection are (2,26,10)
Required line is perpendicular to both ${\bar{n}}_{1} \times {\bar{n}}_{2} & {\bar{n}}_{2}$
Dr’s of the required line are -44,13,-25
Equation of the required line

$\frac{x - 2}{- 44} = \frac{y + 1}{13} = \frac{z - 1}{- 25}$