To find the point $P(x, y, z)$ on the line $\ell$ that is nearest to the origin, we need to minimize the distance from the origin to any point on the line.

The line $\ell$ is given by the parametric equation:

$r(t) = (8, 7, 3) + t(2, 9, 3), \quad t \in \mathbb{R}$

A point on the line can be represented as:

$P(t) = (x(t), y(t), z(t)) = (8 + 2t, 7 + 9t, 3 + 3t)$

The distance $D$ from the origin to the point $P(t)$ is given by:

$D = \sqrt{x(t)^2 + y(t)^2 + z(t)^2}$

To minimize this distance, we minimize the squared distance $D^2$ (since the square root is a monotonic function, the minimum of $D^2$ will also minimize $D$):

$D^2 = (x(t))^2 + (y(t))^2 + (z(t))^2$

Substitute the expressions for $x(t), y(t), z(t)$:

$D^2 = (8 + 2t)^2 + (7 + 9t)^2 + (3 + 3t)^2$

Now, expand and simplify the expression:

$D^2 = (64 + 32t + 4t^2) + (49 + 126t + 81t^2) + (9 + 18t + 9t^2)$

$D^2 = (64 + 49 + 9) + (32t + 126t + 18t) + (4t^2 + 81t^2 + 9t^2)$

$D^2 = 122 + 176t + 94t^2$

To minimize $D^2$, take the derivative with respect to $t$ and set it to zero:

$\frac{d}{dt}(D^2) = 2 \cdot 94t + 176 = 188t + 176$

Set the derivative equal to zero and solve for $t$:

$188t + 176 = 0$

$188t = -176$

$t = -\frac{176}{188} = -\frac{88}{94} = -\frac{44}{47}$

Now substitute $t = -\frac{44}{47}$ back into the parametric equations to find the coordinates $(x, y, z)$:

$x = 8 + 2\left(-\frac{44}{47}\right) = 8 - \frac{88}{47} = \frac{376}{47} - \frac{88}{47} = \frac{288}{47}$

$y = 7 + 9\left(-\frac{44}{47}\right) = 7 - \frac{396}{47} = \frac{329}{47} - \frac{396}{47} = -\frac{67}{47}$

$z = 3 + 3\left(-\frac{44}{47}\right) = 3 - \frac{132}{47} = \frac{141}{47} - \frac{132}{47} = \frac{9}{47}$

So, the coordinates of the point $P(x, y, z)$ that is nearest to the origin are:

$\boxed{\left(\frac{288}{47}, -\frac{67}{47}, \frac{9}{47}\right)}$

Would you like more details, or have any questions?

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Here are some related questions you might find interesting:

1. How do you derive the formula for the distance between a point and a line in 3D space?
2. Can you find the distance from the origin to the line directly using a vector approach?
3. How would the solution change if the line equation was given in a different form?
4. What is the geometric interpretation of minimizing the squared distance in this context?
5. Can you generalize this method to find the closest point on any line to any given point in space?

Tip: Remember that minimizing a function's square often simplifies the algebra, avoiding the complexities of square roots.