It is given that,

A line passes through the point P(3, 0, 1) and parallel to the planes x + 2y = 0 and 3y – z = 0.

We need to find the equation of the line.

Let the position vector of the point P(3, 0, 1) be

Or,

…(i)

Let be the normal vector to the given plane.

Then, is perpendicular to the normal of the plane x + 2y = 0 and 3y – z = 0.

Normal of the plane x + 2y = 0 is

Normal of the plane 3y – z = 0 is

So, is perpendicular to the vectors and .

So,

Take 1^st row and 1^st column, multiply the first element of the row (a₁₁) with the difference of multiplication of opposite elements (a₂₂ × a₃₃ – a₂₃ × a₃₂), excluding 1^st row and 1^st column.

Here,

Now take 1^st row and 2^nd column, multiply the second element of the row (a₁₂) with the difference of multiplication of opposite elements (a₂₁ × a₃₃ – a₂₃ × a₃₁), excluding 1^st row and 2^nd column.

Here,

Similarly, take 1^st row and 3^rd column, multiply the third element of the row (a₁₃) with the difference of multiplication of opposite elements (a₂₂ × a₃₃ – a₂₃ × a₃₂), excluding 1^st row and 3^rd column.

Here,

Further simplifying it,

So, the direction ratio of is (-2, 1, 3). …(ii)

We know that,

Vector equation of a line passing through a point and parallel to a vector is , where λ ∈ ℝ.

Here, from (i) and (ii),

Putting these vectors in , we get

But we know that,

Substituting it,

Thus, the required equation of the line is .