A plane P is parallel to two lines whose direction ratios are $$(-2, 1, -3)$$, and $$(-1, 2, -2)$$ and it contains the point $$(2, 2, -2)$$. Let P intersect the co-ordinate axes at the points A, B, C making the intercepts $$\alpha, \beta, \gamma$$. If V is the volume of the tetrahedron OABC, where O is the origin and $$p = \alpha + \beta + \gamma$$, then the ordered pair $$(V, p)$$ is equal to

We need a plane $$P$$ parallel to both lines with direction ratios $$(-2, 1, -3)$$ and $$(-1, 2, -2)$$, and containing the point $$(2, 2, -2)$$.

The normal to the plane is perpendicular to both direction vectors. We compute the cross product:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & -3 \\ -1 & 2 & -2 \end{vmatrix} = \hat{i}(-2+6) - \hat{j}(4-3) + \hat{k}(-4+1) = (4, -1, -3)$$

The equation of the plane through $$(2, 2, -2)$$ with normal $$(4, -1, -3)$$ is:

$$4(x-2) - 1(y-2) - 3(z+2) = 0$$

$$4x - 8 - y + 2 - 3z - 6 = 0$$

$$4x - y - 3z = 12$$

The intercepts are found by setting two variables to zero each time:

$$x$$-intercept ($$\alpha$$): set $$y = 0, z = 0$$: $$4x = 12$$, so $$\alpha = 3$$.

$$y$$-intercept ($$\beta$$): set $$x = 0, z = 0$$: $$-y = 12$$, so $$\beta = -12$$.

$$z$$-intercept ($$\gamma$$): set $$x = 0, y = 0$$: $$-3z = 12$$, so $$\gamma = -4$$.

Now $$p = \alpha + \beta + \gamma = 3 + (-12) + (-4) = -13$$.

The volume of the tetrahedron $$OABC$$ with vertices at the origin and the intercept points is:

$$V = \frac{|\alpha \cdot \beta \cdot \gamma|}{6} = \frac{|3 \times (-12) \times (-4)|}{6} = \frac{144}{6} = 24$$

Hence, the ordered pair $$(V, p) = (24, -13)$$, and the correct answer is Option B.