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

Dear Student,
Please find below the solution to the asked query:

Given,x+12=y+23=z+34=kany point on this line can be written as:x=2k-1y=3k-2z=4k-3putting it in eqn of plane:x+4y+4z=62k-1+43k-2+44k-3=62k-1+12k-8+16k-12=630k=27k=2730=910x=2k-1=1810-1=45y=3k-2=2710-2=710z=4k-3=3610-3=35hence the point is 45,710,35

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