Find the values of p for which the equation x2 − 3x + 1 = p(x−3) has equal roots.

The given equation is $x^2 - 3x + 1 = p(x-3)$.
Expand the right side:

$$x^2 - 3x + 1 = px - 3p$$

Move all terms to the left side:

$$x^2 - 3x - px + 1 + 3p = 0$$

Factor out $x$:

$$x^2 + (-3 - p)x + (1 + 3p) = 0$$

From the standard quadratic form $ax^2 + bx + c = 0$, we have:

$$a = 1$$ $$b = -3 - p$$ $$c = 1 + 3p$$

For an equation to have equal roots, the discriminant ($\Delta$) must be zero. The discriminant is given by the formula $\Delta = b^2 - 4ac$.
Set the discriminant to zero:

$$b^2 - 4ac = 0$$

Substitute the values of $a$, $b$, and $c$:

$$(-3 - p)^2 - 4(1)(1 + 3p) = 0$$

Simplify the equation:

$$(p + 3)^2 - 4(1 + 3p) = 0$$ $$p^2 + 6p + 9 - 4 - 12p = 0$$

Combine like terms:

$$p^2 - 6p + 5 = 0$$

Factor the quadratic equation:

$$(p - 1)(p - 5) = 0$$

Set each factor to zero to find the possible values of $p$:

$$p - 1 = 0 \implies p = 1$$ $$p - 5 = 0 \implies p = 5$$

The possible values of $p$ are $\boxed{1, 5}$.