Find the coordinates of the point where the line `("x"+1"\ ")/2=("y"+2)/3=("z"+3)/4` meets the plane `"x"+"y"+4"z"=6.`

Explanation

The equation of the given line is
`(x+1)/2=(y+2)/3=(z+3)/4 = lambda` (say)………(i)
The equation of the given plane is
`x+y+4z=6`……………..(i)
The general point on the line (i) is `(2lambda-1,3lambda-2, 4lambda-3)`.
Let the given line (i) meet the given plane (ii) at the point `P(2lambda-1,3lambda-2,4lambda-3)` for some value of `lambda`.
Then, `(2lambda-1)+(3lambda-2)+4(4lambda-3)=6 rArr 21lambda=21 rArr lambda=1`.
Hence, the required point of intersection of the given line (i) and the given plane (ii) is `P(2 xx 1-1,3 xx 1-2, 4 xx1-3)`, i.e., P(1,1,1).