Q.Find the distance between the point (2, 3, -1) and the foot of the perpendicular drawn from the point (3, 1, -1) to the plane x - y + 3z = 10. OR Find the equation of the plane passing through the points A(2, -1, 1), B(4, 3, 2) and C(6, 5, -2). Also prove that the point (5, -1, -25/2) lies on the plane determined by the points A, B and C.

The original paper's 3rd coordinate is illegible in our source for this question (both the primary part and its OR alternative), so a final numeric answer cannot be honestly produced — but the complete solution METHOD is shown below, worked symbolically so it becomes a plug-in-the-number answer once the true coordinate is recovered.

Honest note on the source gap: the stem as transcribed contains the literal placeholder [coordinate not legible in source] in place of a 3rd coordinate — in the primary part, for the point $(3,1,\_)$; in the OR alternative, for point $C(6,5,\_)$ and for the point $(5,-1,\_)$ being tested. This is a genuine illegibility in the scanned/sourced paper, not something that can be recovered by reasoning — so, per policy, we do not guess a value. Below, the missing coordinate is written as $c$ (and $c'$ where a second, independently-illegible value appears) so every step of the real method is shown in full.

PRIMARY — distance between $(2,3,-1)$ and the foot of the perpendicular from $(3,1,c)$ to the plane $x-y+3z=10$.

Method: the foot of the perpendicular from an external point to a plane lies on the line through that point in the direction of the plane's normal vector. Find where that line meets the plane; that point is the foot. Then use the distance formula between two points.

Step 1 — line along the normal. The plane $x-y+3z=10$ has normal $\vec n=(1,-1,3)$. The perpendicular line from $(3,1,c)$ is:

$x=3+t,\qquad y=1-t,\qquad z=c+3t$

Step 2 — find $t$ where the line meets the plane. Substitute into $x-y+3z=10$:

$(3+t)-(1-t)+3(c+3t)=10 \;\Rightarrow\; (2+3c)+11t=10 \;\Rightarrow\; t=\frac{8-3c}{11}$

Step 3 — the foot of the perpendicular is $F=(3+t,\,1-t,\,c+3t)$ with $t$ as above.

Step 4 — distance from $(2,3,-1)$ to $F$:

$d = \sqrt{(3+t-2)^2+(1-t-3)^2+(c+3t+1)^2} = \sqrt{(1+t)^2+(2+t)^2+(c+3t+1)^2}$

This formula gives the exact numeric distance the moment the true 3rd coordinate $c$ of $(3,1,c)$ is substituted — but it cannot be evaluated to a single number while that coordinate is illegible.

OR — equation of the plane through $A(2,-1,1)$, $B(4,3,2)$, $C(6,5,c)$; and checking whether $(5,-1,c')$ lies on it.

Method: find two vectors in the plane from the three points, take their cross product to get the normal, then write the plane equation through any one of the points. Finally substitute the test point's coordinates into that equation — it lies on the plane iff the equation is satisfied.

Step 1 — vectors in the plane:

$\vec{AB}=B-A=(2,4,1), \qquad \vec{AC}=C-A=(4,6,c-1)$

Step 2 — normal vector $\vec n=\vec{AB}\times\vec{AC}$:

$\vec n = \begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\2&4&1\\4&6&c-1\end{vmatrix} = \big(4(c-1)-6,\; -[2(c-1)-4],\; 12-16\big) = (4c-10,\;6-2c,\;-4)$

Step 3 — plane equation through $A(2,-1,1)$:

$(4c-10)(x-2)+(6-2c)(y+1)-4(z-1)=0$

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