Think geometrically !

Consider a plane $3x + 2y + z =7$ , And $(a,b,c)$ be any point lying on it . So we have to find the square of the minimum distance of the plane from the origin.

Which is equal to $(7/√14)^{2}$ = $7/2 = 3.5$

Using Cauchy-Schwarz inequality :

$\begin{aligned} (3a+2b+c)^2 & \le (3^2+2^2+1^2) (a^2+b^2+c^2) \\ 7^2 &\le 14 (a^2+b^2+c^2) \\ \implies a^2+b^2+c^2 & \ge \frac 72 = \boxed{3.5} \end{aligned}$

Let i,j,k be three Non coplanar unit vectors along x,y,z axis.

Consider two vectors

A = ai+bj+ck

B = 3i+2j+k

Taking dot product

A.B = 3a+2b+c = 7

We know that A.B = mod(A).mod(B).cos(x)

where x is angle between A And B

We need to minimise mod B hence we take cosx = 1

Doing this we get required answer

Using Lagrange Multipliers, let $f(a,b,c) = a^2 + b^2 + c^2$ and $g(a,b,c) = 3a + 2b + c = 7$ such that $grad(f) = \lambda \cdot grad(g).$ This yields the system:

$2a = \lambda \cdot 3;$

$2b = \lambda \cdot 2;$

$2c = \lambda \cdot 1;$

or $b = \frac{2a}{3}, c = \frac{c}{3} \Rightarrow 3a + 2(\frac{2a}{3}) + \frac{a}{3} = 7 \Rightarrow a = \frac{3}{2}, b = 1, c =\frac{1}{2}.$

Now the Hessian matrix of $f(a,b,c)$ , $F(a,b,c)$ , is just $F = 2 \cdot I_{3x3}$ (where $I_{3x3}$ is the 3x3 identity matrix), which is positive-definite for all $(a,b,c)$ . Hence the minimum value is $f(\frac{3}{2}, 1, \frac{1}{2}) = (\frac{3}{2})^2 + 1^2 + (\frac{1}{2})^2 = \boxed{\frac{7}{2}}.$