The equation of line of intersection of the planes x + y + 3z = 7 and…

The correct answer is Option (4) → \(\displaystyle\frac{x-5}{5} = \frac{y-2}{1} = \frac{z}{-2}\)

The line lies in both planes, so it is perpendicular to both normals (1, 1, 3) and (1, −1, 2). Its direction is the cross product of the normals:

    \( \displaystyle \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 1 & 3 \\ 1 & -1 & 2 \end{vmatrix} = \hat{\imath}(2+3) - \hat{\jmath}(2-3) + \hat{k}(-1-1) \)

    \( = 5\hat{\imath} + \hat{\jmath} - 2\hat{k} \)

So the direction ratios are 5, 1, −2. For a point on the line, put z = 0:

    \(x + y = 7\)
    \(x - y = 3\)
    \(x = 5,\ y = 2\)

The line through (5, 2, 0) with direction ratios 5, 1, −2 is

    \( \displaystyle \frac{x-5}{5} = \frac{y-2}{1} = \frac{z}{-2} \)

Note: This topic was not in the CUET 2026 syllabus.