Question
The equation of the line passing through $$\left( { - 4,\,3,\,1} \right),$$ parallel to the plane $$x+2y-z-5=0$$ and intersecting the line $$\frac{{x + 1}}{{ - 3}} = \frac{{y - 3}}{2} = \frac{{z - 2}}{{ - 1}}$$ is :
A.
$$\frac{{x - 4}}{2} = \frac{{y + 3}}{1} = \frac{{z + 1}}{4}$$
B.
$$\frac{{x + 4}}{1} = \frac{{y - 3}}{1} = \frac{{z - 1}}{3}$$
C.
$$\frac{{x + 4}}{3} = \frac{{y - 3}}{{ - 1}} = \frac{{z - 1}}{1}$$
D.
$$\frac{{x + 4}}{{ - 1}} = \frac{{y - 3}}{1} = \frac{{z - 1}}{1}$$
Answer :
$$\frac{{x + 4}}{3} = \frac{{y - 3}}{{ - 1}} = \frac{{z - 1}}{1}$$
Solution :
Let any point on the intersecting line
$$\frac{{x + 1}}{{ - 3}} = \frac{{y - 3}}{2} = \frac{{z - 2}}{{ - 1}} = \lambda \,\,\,\,\left( {{\text{say}}} \right)$$
is $$\left( { - 3\lambda - 1,\,2\lambda + 3,\, - \lambda + 2} \right)$$
Since, the above point lies on a line which passes through the point $$\left( { - 4,\,3,\,1} \right)$$
Then, direction ratio of the required line
$$\eqalign{
& = \left\langle { - 3\lambda - 1 + 4,\,2\lambda + 3 - 3,\, - \lambda + 2 - 1} \right\rangle \cr
& {\text{or,}}\,\left\langle { - 3\lambda + 3,\,2\lambda ,\, - \lambda + 1} \right\rangle \cr} $$
Since, line is parallel to the plane $$x+2y-z-5=0$$
Then, perpendicular vector to the line is $$\hat i + 2\hat j - \hat k$$
$$\eqalign{
& {\text{Now }}\left( { - 3\lambda + 3} \right)\left( a \right) + \left( {2\lambda } \right)\left( b \right) + \left( { - \lambda + 1} \right)\left( { - 1} \right) = 0 \cr
& \Rightarrow \lambda = - 1 \cr} $$
Now direction ratio of the required line
$$ = \left\langle {6,\, - 2,\,2} \right\rangle {\text{ or }}\left\langle {3,\, - 1,\,1} \right\rangle $$
Hence required equation of the line is $$\frac{{x + 4}}{3} = \frac{{y - 3}}{{ - 1}} = \frac{{z - 1}}{1}$$