If `P_(1) and P_(2)` are the perpendicular of the points with position vectors `veca = 3veci - 5 hatj+8hatk` and
`vecb = 2veci - 41 hatj+ 21hatk` from the plane `vecr cdot (2hati + 3hatj -hatk)=12, `then `P_(1)^(2)+P_(2)^(2)`is equal to __________.

Answer

To solve the problem, we need to find the perpendicular distances from the points with position vectors \(\vec{a} = 3\hat{i} - 5\hat{j} + 8\hat{k}\) and \(\vec{b} = 2\hat{i} - 41\hat{j} + 21\hat{k}\) to the plane given by the equation \(\vec{r} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 12\).

### Step 1: Write the equation of the plane in standard form
The equation of the plane can be expressed as:
\[ 2x + 3y - z = 12 \]

### Step 2: Identify the normal vector of the plane
The normal vector \(\vec{n}\) of the plane is given by the coefficients of \(x\), \(y\), and \(z\) in the plane equation:
\[ \vec{n} = 2\hat{i} + 3\hat{j} - \hat{k} \]

### Step 3: Calculate the perpendicular distance \(P_1\) from point \(\vec{a}\) to the plane
Using the formula for the distance \(d\) from a point \((x_0, y_0, z_0)\) to the plane \(Ax + By + Cz + D = 0\):
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
For point \(\vec{a} = (3, -5, 8)\):
- \(A = 2\), \(B = 3\), \(C = -1\), \(D = -12\)
- \(x_0 = 3\), \(y_0 = -5\), \(z_0 = 8\)

Substituting these values:
\[ P_1 = \frac{|2(3) + 3(-5) - 8 - 12|}{\sqrt{2^2 + 3^2 + (-1)^2}} = \frac{|6 - 15 - 8 - 12|}{\sqrt{4 + 9 + 1}} = \frac{|-29|}{\sqrt{14}} = \frac{29}{\sqrt{14}} \]

### Step 4: Calculate the perpendicular distance \(P_2\) from point \(\vec{b}\) to the plane
For point \(\vec{b} = (2, -41, 21)\):
\[ P_2 = \frac{|2(2) + 3(-41) - 21 - 12|}{\sqrt{2^2 + 3^2 + (-1)^2}} = \frac{|4 - 123 - 21 - 12|}{\sqrt{14}} = \frac{|-152|}{\sqrt{14}} = \frac{152}{\sqrt{14}} \]

### Step 5: Calculate \(P_1^2 + P_2^2\)
Now we need to find \(P_1^2 + P_2^2\):
\[ P_1^2 = \left(\frac{29}{\sqrt{14}}\right)^2 = \frac{841}{14} \]
\[ P_2^2 = \left(\frac{152}{\sqrt{14}}\right)^2 = \frac{23104}{14} \]
Adding these together:
\[ P_1^2 + P_2^2 = \frac{841 + 23104}{14} = \frac{23945}{14} \]

### Final Answer
Thus, the value of \(P_1^2 + P_2^2\) is:
\[ \frac{23945}{14} \]