Let a line pass through two distinct points P(−2,−1, 3) and , and be parallel to the vector $$3\widehat{i}+2\widehat{j}+2\widehat{k}$$. If the distance of the point Q from the point R(1, 3, 3) is 5 , then the square of the area of △PQR is equal to :

The line passes through point P(-2, -1, 3) and is parallel to the vector $$3\widehat{i} + 2\widehat{j} + 2\widehat{k}$$. Therefore, the parametric equations for any point on the line are:

$$x = -2 + 3t$$
$$y = -1 + 2t$$
$$z = 3 + 2t$$

where $$t$$ is a scalar parameter.

Let Q be a point on this line, so Q has coordinates $$(-2 + 3t, -1 + 2t, 3 + 2t)$$. The distance from Q to R(1, 3, 3) is given as 5. Using the distance formula:

$$\sqrt{ [(-2 + 3t) - 1]^2 + [(-1 + 2t) - 3]^2 + [(3 + 2t) - 3]^2 } = 5$$

Simplify the expressions inside the square root:

$$(-2 + 3t) - 1 = 3t - 3$$
$$(-1 + 2t) - 3 = 2t - 4$$
$$(3 + 2t) - 3 = 2t$$

So the equation becomes:

$$\sqrt{ (3t - 3)^2 + (2t - 4)^2 + (2t)^2 } = 5$$

Square both sides to eliminate the square root:

$$(3t - 3)^2 + (2t - 4)^2 + (2t)^2 = 25$$

Expand each term:

$$(3t - 3)^2 = 9t^2 - 18t + 9$$
$$(2t - 4)^2 = 4t^2 - 16t + 16$$
$$(2t)^2 = 4t^2$$

Sum them up:

$$9t^2 - 18t + 9 + 4t^2 - 16t + 16 + 4t^2 = 25$$
$$(9t^2 + 4t^2 + 4t^2) + (-18t - 16t) + (9 + 16) = 25$$
$$17t^2 - 34t + 25 = 25$$

Subtract 25 from both sides:

$$17t^2 - 34t = 0$$

Factor out 17t:

$$17t(t - 2) = 0$$

So, $$t = 0$$ or $$t = 2$$.

When $$t = 0$$, Q is $$(-2 + 3(0), -1 + 2(0), 3 + 2(0)) = (-2, -1, 3)$$, which is point P. However, if Q coincides with P, the triangle PQR degenerates to a line, so we discard this solution.

When $$t = 2$$, Q is $$(-2 + 3(2), -1 + 2(2), 3 + 2(2)) = (4, 3, 7)$$.

Thus, the points are P(-2, -1, 3), Q(4, 3, 7), and R(1, 3, 3). To find the area of triangle PQR, use the formula for the area of a triangle given three points in space. The area is half the magnitude of the cross product of vectors $$\overrightarrow{PQ}$$ and $$\overrightarrow{PR}$$.

First, find the vectors:

$$\overrightarrow{PQ} = Q - P = (4 - (-2), (3 - (-1)), (7 - 3)) = (6, 4, 4)$$
$$\overrightarrow{PR} = R - P = (1 - (-2)), (3 - (-1)), (3 - 3)) = (3, 4, 0)$$

Now, compute the cross product $$\overrightarrow{PQ} \times \overrightarrow{PR}$$:

$$\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 6 & 4 & 4 \\ 3 & 4 & 0 \\ \end{vmatrix} = \widehat{i}(4 \cdot 0 - 4 \cdot 4) - \widehat{j}(6 \cdot 0 - 4 \cdot 3) + \widehat{k}(6 \cdot 4 - 4 \cdot 3)$$

Calculate each component:

$$\widehat{i}(0 - 16) = -16\widehat{i}$$
$$-\widehat{j}(0 - 12) = -\widehat{j}(-12) = 12\widehat{j}$$
$$\widehat{k}(24 - 12) = 12\widehat{k}$$

So, the cross product vector is $$(-16, 12, 12)$$.

The magnitude of this vector is:

$$\sqrt{(-16)^2 + 12^2 + 12^2} = \sqrt{256 + 144 + 144} = \sqrt{544}$$

Simplify $$\sqrt{544}$$:

$$544 = 16 \times 34$$
$$\sqrt{544} = \sqrt{16 \times 34} = 4\sqrt{34}$$

Thus, the area of triangle PQR is:

$$\frac{1}{2} \times 4\sqrt{34} = 2\sqrt{34}$$

The square of the area is:

$$(2\sqrt{34})^2 = 4 \times 34 = 136$$

Therefore, the square of the area of $$\triangle PQR$$ is 136.