If the point of intersection of the lines \(\frac{x+1}{3}=\frac{y+a}{5}=\frac{z+b+1}{7}\) and \(\frac{x-2}{1}=\frac{y-b}{4}=\frac{z-2a}{7}\) lies on \(xy-\)plane, then the value of \(a+b\) is:

[JEE Main 2026, 2 Apr (Shift 1)]

Question

If the point of intersection of the lines \(\frac{x+1}{3}=\frac{y+a}{5}=\frac{z+b+1}{7}\) and \(\frac{x-2}{1}=\frac{y-b}{4}=\frac{z-2a}{7}\) lies on \(xy-\)plane, then the value of \(a+b\) is:

[JEE Main 2026, 2 Apr (Shift 1)]

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If the point of intersection of the lines \(\frac{x+1}{3}=\frac{y+a}{5}=\frac{z+b+1}{7}\) and \(\frac{x-2}{1}=\frac{y-b}…

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