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

Answer» Let `(x+1)/2= (y+2)/3 = (z+3)/4 = k`
Then, `x = 2k-1, y = 3k-2, z = 4k-3`
Putting these values in the equation of given plane `x+y+4z = 6`,
`2k-1+3k-2+16k-12 = 6`
`->21k = 21=> k = 1`
So, coordinates will be `(2(1)-1,3(1)-2,4(1)-3) = (1,1,1)`.


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